All Solutions
Page 611: Questions
5(x – 7) = 5x – 35
Check by giving x any value; 5 for example
5(5 – 7) = 5 $times$ 5 – 35
5 $times$ -2 = 25 – 35
-10 = -10
So, it is always true
2x$^{2}$ = 50
This is sometimes true
Find possible values for x:
x$^{2} = dfrac{50}{2} = 25$
X = $sqrt{25} = pm$ 5
So, it is true only when x = 5 or – 5
4x – (3x + 2) = x – 7
4x – 3x – 2 = x- 7
x – 2 = x – 7
Same coefficient for x on both sides
So, this is never true
$dfrac{2}{3}$(x – 9) = $dfrac{3}{4}$(2x + 5)
This is sometimes true
Find possible values for x
8x – 72 = 18x + 45
x = – 11.7
So, it is true only when x = – 11.7
b) sometimes true
c) never true
d) sometimes true
{color{#4257b2}text{a)}}
$$
Solution to this example is given below
$$
begin{align*}
5x-35&=5x-35&&boxed{text{Distributive property}}\
5x-35+35&=5x-35+35&&boxed{text{Add }35 text{ to both sides}}\
5x&=5x&&boxed{text{Simplify}}\
5x-5x&=5x-5x&&boxed{text{Subtract } 5x text{ from both sides}}\
0&=0&&boxed{text{Simplify}}\
end{align*}
$$
Both sides are equal. The equation is always true.
$$
begin{align*}
&boxed{{color{#c34632}text{True for all } x} }&&boxed{text{Final solution}}\
end{align*}
$$
{color{#4257b2}text{b)}}
$$
Solution to this example is given below
$$
begin{align*}
frac{2x^2}{2}&=frac{50}{2}&&boxed{text{Divide both sides by }2}\
x^2&=25&&boxed{text{Simplify}}\
x=sqrt{25},:x&=-sqrt{25}&&boxed{text{Simplify}}\
end{align*}
$$
First we solve $sqrt{25}$
$$
begin{align*}
&sqrt{5^2}&&boxed{text{Factor the number: } 25=5^2}\
&5&&boxed{text{Apply radical rule}}\
end{align*}
$$
Second we solve $-sqrt{25}$
$$
begin{align*}
&-sqrt{5^2}&&boxed{text{Factor the number: } 25=5^2}\
&-5&&boxed{text{Apply radical rule}}\
end{align*}
$$
So, it is true only when $boldsymbol{x=5}, text { or } boldsymbol{x=-5}$
$$
begin{align*}
&boxed{{color{#c34632}text{Sometimes true}} }&&boxed{text{Final solution}}\
end{align*}
$$
$$
boxed{ color{#c34632} text{ } mathrm{For:}x^2=fleft(aright)mathrm{:the:solutions:are:}x=sqrt{fleft(aright)},::-sqrt{fleft(aright)}}
$$
$$
boxed{ color{#c34632} text{ } mathrm{Apply:radical:rule}:quad sqrt[n]{a^n}=a}
$$
$$
boxed{ color{#c34632} text{ }mathrm{Apply:radical:rule}:quad sqrt[n]{a^n}=a }
$$
{color{#4257b2}text{c)}}
$$
Solution to this example is given below
$$
begin{align*}
4x-3x-2&=x-7&&boxed{text{Distributive property}}\
x-2&=x-7&&boxed{text{Combine like terms: }4x-3x=x}\
x-2+2&=x-7+2&&boxed{text{Add 2 to both sides}}\
x&=x-5&&boxed{text{Simplify}}\
x-x&=x-5-x&&boxed{text{Subtract } 5x text{ from both sides}}\
0&=-5&&boxed{text{Simplify}}\
end{align*}
$$
Both sides are not equal.
Same coefficient for $x$ on both sides
The equation is never true.
$$
begin{align*}
&boxed{{color{#c34632}text{Never True } } }&&boxed{text{Final solution}}\
end{align*}
$$
{color{#4257b2}text{d)}}
$$
Solution to this example is given below
$$
begin{align*}
frac{2}{3}x-6&=frac{3}{2}x+frac{15}{4}&&boxed{text{Distributive property}}\
frac{2}{3}x-6+6&=frac{3}{2}x+frac{15}{4}+6&&boxed{text{Add }6 text{ to both sides}}\
frac{2}{3}x&=frac{39}{4}+frac{3x}{2}&&boxed{text{Simplify}}\
frac{2}{3}x-frac{3x}{2}&=frac{39}{4}+frac{3x}{2}-frac{3x}{2}&&boxed{text{Subtract } frac{3x}{2} text{ from both sides}}\
-frac{5x}{6}&=frac{39}{4}&&boxed{text{Simplify}}\
6left(-frac{5x}{6}right)&=frac{39cdot :6}{4}&&boxed{text{Multiply both sides by 6}}\
-5x&=frac{117}{2}&&boxed{text{Simplify}}\
frac{-5x}{-5}&=frac{frac{117}{2}}{-5}&&boxed{text{Divide both sides by } -5}\
x&=-frac{117}{10}&&boxed{text{Simplify}}\
x&=-11.7&&boxed{text{Convert to a decimal form}}\
end{align*}
$$
So, it is true only when $boldsymbol{x=-11.7}$
$$
begin{align*}
&boxed{{color{#c34632}text{Sometimes true}} }&&boxed{text{Final solution}}\
end{align*}
$$
color{#4257b2} text{ a) True for all }x
$$
$$
color{#4257b2} text{ b) Sometimes true}
$$
$$
color{#4257b2} text{ c) Never true}
$$
$$
color{#4257b2} text{ d) Sometimes true }
$$
sin x=dfrac{1}{2}
$$
sin x=4
$$
cos x=2
$$
cos x=-dfrac{1}{2}
$$
sin x=cosleft(dfrac{pi}{2}-xright)
$$
sin x=cosleft(dfrac{pi}{2}-xright)+3
$$
sin x=sinleft(dfrac{pi}{2}-xright)
$$
$2sin dfrac{x-dfrac{pi}{2}+x}{2}cosdfrac{x+dfrac{pi}{2}-x}{2}=0$
$2sinleft(x-dfrac{pi}{4}right)cos dfrac{pi}{4}=0$
$sinleft(x-dfrac{pi}{4}right)=0$
$x=dfrac{pi}{4}+kpi, k$ integer
sin x=sinleft(dfrac{pi}{2}-xright)+6
$$
tan x=0
$$
tandfrac{7pi}{2}+x=0
$$
$$
sin x=dfrac{sqrt 3}{2}
$$
2sin x=-7
$$
sin x=dfrac{3}{2}
$$
sin x=dfrac{2}{3}
$$
cos x=sinleft(dfrac{pi}{2}-xright)
$$
cos x=sinleft(dfrac{pi}{2}-xright)-5
$$
It means that irrespective of the value of $x$, the output or the solution will not change. In other words we can say that the solution of the given equation does not depend on the value of $x$.
$textbf{b.}$ Sometimes true.
It means that the value of $x$ affect the solution of the given equation,
for example., $sin x = dfrac{1}{2}$
the solution for the above equation will be of $4n+1$, where $n=0,1,2,…$
$textbf{c.}$ Never true.
Irrespective of the value of $x$, the solution would never be true.
For example, $sin x= 2$
For any value of $x$ we will not get the solution.
$textbf{b.}$ It means that the value of $x$ affect the solution of the given equation,
$textbf{c.}$ Irrespective of the value of $x$, the solution would never be true.
{color{#4257b2}text{a)}}
$$
Solution to this example is given below
$$
begin{align*}
3x-2-4x-4&=-x-6&&boxed{text{Distributive property}}\
-x-2-4&=-x-6&&boxed{text{Combine like terms: }3x-4x=-x}\
-x-6&=-x-6&&boxed{text{Subtract the numbers: }-2-4=-6}\
-x-6+6&=-x-6+6&&boxed{text{Add } 6 text{ to both sides}}\
-x&=-x&&boxed{text{Simplify}}\
-x+x&=-x+x&&boxed{text{Add } xtext{ to both sides}}\
0&=0&&boxed{text{Simplify}}\
end{align*}
$$
Both sides are equal. The equation is always true
$$
begin{align*}
&boxed{{color{#c34632}text{True for all }x } }&&boxed{text{Final solution}}\
end{align*}
$$
{color{#4257b2}text{b)}}
$$
Solution to this example is given below
$$
begin{align*}
3x-5&=2x+2+x&&boxed{text{Distributive property}}\
3x-5&=3x+2&&boxed{text{Combine like terms: }2x+x=3x}\
3x-5+5&=3x+2+5&&boxed{text{Add 5 to both sides}}\
3x&=3x+7&&boxed{text{Simplify}}\
3x-3x&=3x+7-3x&&boxed{text{Subtract } 3xtext{ from both sides}}\
0&=7&&boxed{text{Simplify}}\
end{align*}
$$
Both sides are not equal. The equation is never true.
$$
begin{align*}
&boxed{{color{#c34632}text{No solution} } }&&boxed{text{Final solution}}\
end{align*}
$$
{color{#4257b2}text{c)}}
$$
Solution to this example is given below
Manipulating right side
$$
begin{align*}
&cos left(frac{pi }{2}-xright)&&boxed{text{Simplify}}\
&cos left(frac{pi }{2}right)cos left(xright)+sin left(frac{pi }{2}right)sin left(xright)&&boxed{text{Use the following identity}}\
&0+sin left(xright)&&boxed{text{Simplify}}\
&sin(x)&&boxed{text{Simplify}}\
end{align*}
$$
Both sides are equal. The equation is always true.We showed that the
two sides could take the same form.
$$
begin{align*}
&boxed{{color{#c34632}text{True} } }&&boxed{text{Final solution}}\
end{align*}
$$
{color{#4257b2}text{d)}}
$$
Solution to this example is given below
Let’s put $x=180^{circ}$
$$
begin{align*}
&tanpi=0&&boxed{text{Not True}}\\
end{align*}
$$
Then put $x=45^{circ}$
$$
begin{align*}
&tanpi=1&&boxed{text{True}}\
end{align*}
$$
So, It is sometimes true, or sometimes not
color{#4257b2} text{ a) True for all }x
$$
$$
color{#4257b2} text{ b) No solution }
$$
$$
color{#4257b2} text{ c) True }
$$
{color{#4257b2}text{a)}}
$$
Solution to this example is given below
$$
begin{align*}
&x^2-2^2&&boxed{text{Rewrite 4 as }2^2}\
&left(x+2right)left(x-2right)&&boxed{text{Squares formula}}\\
&boxed{{color{#c34632}left(x+2right)left(x-2right)} }&&boxed{text{Final solution}}\
end{align*}
$$
$$
boxed{ color{#c34632} text{ }mathrm{Apply:Difference:of:Two:Squares:Formula::}x^2-y^2=left(x+yright)left(x-yright) }
$$
{color{#4257b2}text{b)}}
$$
Solution to this example is given below
$$
begin{align*}
&y^2-9^2&&boxed{text{Rewrite 81 as }9^2}\
&left(y+9right)left(y-9right)&&boxed{text{Squares formula}}\\
&boxed{{color{#c34632}left(y+9right)left(y-9right)} }&&boxed{text{Final solution}}\
end{align*}
$$
$$
boxed{ color{#c34632} text{ }mathrm{Apply:Difference:of:Two:Squares:Formula::}x^2-y^2=left(x+yright)left(x-yright) }
$$
{color{#4257b2}text{c)}}
$$
Solution to this example is given below
$$
begin{align*}
&-left(x^2-1right)&&boxed{text{Factor out common term } -1}\
&-x^2-1^2&&boxed{text{Rewrite 1 as }1^2}\
&-left(x+1right)left(x-1right)&&boxed{text{Squares formula}}\\
&boxed{{color{#c34632}-left(x+1right)left(x-1right)} }&&boxed{text{Final solution}}\
end{align*}
$$
$$
boxed{ color{#c34632} text{ }mathrm{Apply:Difference:of:Two:Squares:Formula::}x^2-y^2=left(x+yright)left(x-yright) }
$$
{color{#4257b2}text{d)}}
$$
Solution to this example is given below
$$
begin{align*}
&1^2-sin^2(x)&&boxed{text{Rewrite 1 as }1^2}\
&left(1+sin xright)left(1-sin xright)&&boxed{text{Squares formula}}\\
&boxed{{color{#c34632}left(1+sin x right)left(1-sin x right)} }&&boxed{text{Final solution}}\
end{align*}
$$
$$
boxed{ color{#c34632} text{ }mathrm{Apply:Difference:of:Two:Squares:Formula::}x^2-y^2=left(x+yright)left(x-yright) }
$$
color{#4257b2} text{ a) } (x+2)(x-2)
$$
$$
color{#4257b2} text{ b) }(y+9)(y-9)
$$
$$
color{#4257b2} text{ c) }-(x+1)(x-1)
$$
$$
color{#4257b2} text{ d) }(1+sin x)(1-sin x)
$$
According to the formula
$$
color{#c34632}{ sin (2k) = 2 sin x cos x}
$$
we have
$$
sin (2x) = sin ( 2 k pi)=2 sin (k pi) cos (k pi)
$$
When $k$ is a even number, we get
$$
cos ( k pi) = 1
$$
then
$$
begin{align*}
sin ( 2k pi) &= 2 sin (k pi)\
sin (2x) &= 2 sin x
end{align*}
$$
When $k$ is $a$ odd number, we get
$$
cos (k pi) = -1
$$
then
$$
sin (2 k pi) = – 2 sin (k pi)
$$
Since
$$
sin (k pi) = 0
$$
and
$$
sin (2k pi) = 0
$$
we obtain
$$
begin{align*}
sin (2k pi) &= 2 sin (k pi) cos (k pi)\
0 &= 2 cdot 0 cdot ( – 1)\
0 &= 0
end{align*}
$$
Therefore
$$
sin (2x) = 2 sin x
$$
sin (2x) = 2 sin x
$$
$x=kpi, k$ integer
sin (2x)=sin (2kpi)=2sin (kpi)cos (kpi)
$$
$sin (2x)=2sin xcos x$.
sin (2x)=2sin (kpi)cos (kpi)=2cdot 0cos cos (kpi)=textcolor{#4257b2}{0}
$$
2sin x=2sin (kpi)=2cdot 0=textcolor{#4257b2}{0}
$$
f(x)=sin x
$$
Its range is $[-1,1]$.
y=sin 0=0
$$
It has no asymptotes.
It is decreasing on the intervals $left[dfrac{(4k+1)pi}{2},dfrac{(4k+3)pi}{2}right]$ and increasing on the intervals $left[dfrac{(4k-1)pi}{2},dfrac{(4k+1)pi}{2}right]$.
$$
h(x)=cos left(x+dfrac{3pi}{2}right)
$$
$dfrac{3}{left(x-4right)left(x+1right)}+dfrac{6}{x+1}$ (Given)
$=dfrac{3}{left(x-4right)left(x+1right)}+dfrac{6left(x-4right)}{left(x+1right)left(x-4right)}$ (Multiply by the form of 1 unifying denominators.)
$=dfrac{3+6left(x-4right)}{left(x-4right)left(x+1right)}$ (Adding fractions with similar denominators)
$$
=dfrac{6x-21}{left(x-4right)left(x+1right)}
$$
$dfrac{5}{2left(x-5right)}+dfrac{3x}{x-5}$ (Given)
$=dfrac {5}{2(x-5)}+dfrac {3x}{x-5} cdot dfrac {2}{2}$ (Multiply by the form of 1 unifying denominators.)
$=dfrac {6x+5}{2(x-5)}$
$dfrac{x}{x^2-x-2}-dfrac{2}{x^2-x-2}$ (Given)
$=dfrac{x-2}{x^2-x-2}$ (Subtracting fractions with similar denominators)
$=dfrac{x-2}{left(x+1right)left(x-2right)}$ (Factoring polynomial)
$=dfrac{1}{x+1}$ (Simplify)
$dfrac{x+2}{x^2-9}-dfrac{1}{x+3}$ (Given)
$=dfrac{x+2}{left(x+3right)left(x-3right)}-dfrac{1}{x+3}$ (Factoring polynomial)
$=dfrac{x+2}{left(x+3right)left(x-3right)}-dfrac{x-3}{left(x+3right)left(x-3right)}$ (Multiply by the form of 1 unifying denominators.)
$=dfrac{x+2-left(x-3right)}{left(x+3right)left(x-3right)}$ (Subtracting fractions with similar denominators)
$=dfrac{5}{left(x+3right)left(x-3right)}$ (Simplify)
b- $dfrac{5}{2left(x-5right)}+dfrac{3x}{x-5}=dfrac {6x+5}{2(x-5)}$
c- $dfrac{x}{x^2-x-2}-dfrac{2}{x^2-x-2}=dfrac{1}{x+1}$
d- $dfrac{x+2}{x^2-9}-dfrac{1}{x+3}=dfrac{5}{left(x+3right)left(x-3right)}$
begin{cases}
2x+y-3z=-12\
5x-y+z=11\
x+3y-2z=-13
end{cases}
$$
begin{cases}
2x+y-3z+5x-y+z=-12+11\
x+3y-2z+15x-3y+3z=-13+3(11)
end{cases}
$$
$$
begin{cases}
7x-2z=-1\
16x+z=20
end{cases}
$$
$32x+2z+7x-2z=39$
$39x=39$
$$
textcolor{#4257b2}{x=1}
$$
$z=20-16(1)$
$$
textcolor{#4257b2}{z=4}
$$
$5(1)-y+4=11$
$9-y=11$
$y=9-11$
$$
textcolor{#4257b2}{y=-2}
$$
$2(1)+(-2)-3(4)stackrel{?}{=}-12$
$2-2-12stackrel{?}{=}-12$
$-12=-12textcolor{#c34632}{checkmark}$
$5x-y+z=11$
$5(1)-(-2)+4stackrel{?}{=}11$
$5+2+4stackrel{?}{=}11$
$11=11textcolor{#c34632}{checkmark}$
$x+3y-2z=-13$
$1+3(-2)-2(4)stackrel{?}{=}-13$
$1-6-8stackrel{?}{=}-13$
$-13=-13textcolor{#c34632}{checkmark}$
$y=-2$
$$
z=4
$$
$$
begin{align*}
y &= – sqrt {x + 2} + 5\
y &= sqrt {x-3}
end{align*}
$$
Then we have
$$
sqrt {x-3} = – sqrt {x + 2} + 5
$$
We will squares both sides of the equation
$$
begin{align*}
left( sqrt {x-3} right)^2 &= left( – sqrt {x+2} + 5 right)^2\
x-3 &= x+2 – 10 sqrt{x+2} + 25\
10 sqrt{x+2} &= 30\
sqrt{x+2} &= 3
end{align*}
$$
Again, we will squares both sides of the equation
$$
begin{align*}
left( sqrt {x+2} right)^2 &= 3^2\
x+2 &= 9\
x &= 7
end{align*}
$$
Now we can substitute $x=7$
$$
begin{align*}
y &= sqrt {7-3}\
y &= sqrt 4\
y &= 2
end{align*}
$$
The point is
$$
(x,y) = (7,2)
$$
(x,y) = (7,2)
$$
begin{cases}
y=-sqrt{x+2}+5\
y=sqrt{x-3}
end{cases}
$$
-sqrt{x+2}+5=sqrt{x-3}
$$
$x+2+25-10sqrt{x+2}=x-3$
$x+27-10sqrt{x+2}=x-3$
$$
-10sqrt{x+2}=-30
$$
$x+2=9$
$x=9-2$
$$
textcolor{#4257b2}{x=7}
$$
$$
textcolor{#4257b2}{y=2}
$$
$2stackrel{?}{=}-sqrt{7+2}+5$
$2stackrel{?}{=}-sqrt 9+5$
$2stackrel{?}{=}-3+5$
$2=2textcolor{#c34632}{checkmark}$
$y=sqrt{x-3}$
$2stackrel{?}{=}sqrt{7-3}$
$2stackrel{?}{=}sqrt 4$
$2stackrel{?}{=}-3+5$
$2=2textcolor{#c34632}{checkmark}$
$2=2textcolor{#c34632}{checkmark}$
$y=2$
We know that
$$
cos(x-dfrac{pi}{2}) = sin x
$$
Since both function are having period $2pi$. So we can just add $2pi$ get possible values of $h$.
Hence, $h = dfrac{pi}{2}, dfrac{5pi}{2}, dfrac{9pi}{2}$.
$$
sin theta < 0
$$
$$
cos theta < 0
$$
$$
color{#c34632}{ tan theta = frac {sin theta}{cos theta}}
$$
Since $sin theta < 0$ and $cos theta <0$ we obtain
$$
tan theta = frac {<0}{ 0
$$
$$
frac {1}{cos theta} < 0
$$
b) $cos theta 0$
d) $frac 1{cos theta} <0$
f(theta)=sintheta,thetainleft[pi,dfrac{3pi}{2}right]
$$
sinthetaleq 0
$$
f(theta)=costheta,thetainleft[pi,dfrac{3pi}{2}right]
$$
costhetaleq 0
$$
f(theta)=tantheta,thetainleft[pi,dfrac{3pi}{2}right)
$$
tanthetageq 0
$$
The period of function $y= tan x$ is $pi$
.
y=tan x
$$
begin{tabular}{|| c| c||}
hline
$x$ & $y=tan x$ \ [0.5ex]
hline$-dfrac{pi}{3}$ & -1.73 \
hline
$-dfrac{pi}{4}$ & -1 \
hline
0 & 0 \
hline
$dfrac{pi}{4}$ & 1 \
hline
$dfrac{pi}{3}$ & 1.73 \
hline
$dfrac{2pi}{3}$ &-1.73 \
hline
$dfrac{3pi}{4}$ & -1 \
hline
$pi$ & 0 \
hline
$dfrac{5pi}{4}$ & 1 \
hline
$dfrac{4pi}{3}$ & 1.73 \[1ex]
hline
end{tabular}
end{center}
We build a table of values:
$x=dfrac{pi}{2}$
$x=-dfrac{3pi}{2}$
y=tan(x-pi)
$$
begin{tabular}{|| c| c||}
hline
$x$ & $y=tan (x-pi)$ \ [0.5ex]
hline$-dfrac{pi}{3}$ & -1.73 \
hline
$-dfrac{pi}{4}$ & -1 \
hline
0 & 0 \
hline
$dfrac{pi}{4}$ & 1 \
hline
$dfrac{pi}{3}$ & 1.73 \
hline
$dfrac{2pi}{3}$ &-1.73 \
hline
$dfrac{3pi}{4}$ & -1 \
hline
$pi$ & 0 \
hline
$dfrac{5pi}{4}$ & 1 \
hline
$dfrac{4pi}{3}$ & 1.73 \[1ex]
hline
end{tabular}
end{center}
We build a table of values:
$x=dfrac{pi}{2}$
$x=-dfrac{3pi}{2}$
$$
6x-3y+9z = 18
$$
b) On the graph is shown the plane $6x-3y+9z = 18$
P(2,-1,-3)
$$
6x-3y+9z=18
$$
$x=0;z=0Rightarrow -3y=18Rightarrow y=-6$
$x=0;y=0Rightarrow 9z=18Rightarrow z=2$
$$
begin{align*}
frac {x}{1- frac 1x} &= frac {x}{frac {x-1}{x}} && text{$left(1- frac 1x=frac xx-frac 1x=frac {x-1}{x} right)$}\
&= frac {x cdot x}{x-1} && text{$left(frac a{frac bc}=frac {a cdot c}b right)$}\
&= frac {x^2}{x-1}
end{align*}
$$
$$
frac {x}{1- frac 1x} = frac {x^2}{x-1}
$$
$$
begin{align*}
frac {frac 1a + frac 1b}{frac 1b – a} &= frac {frac {b+a}{ab}}{frac {1-ab}{b}} && text{$left( frac 1b – a=frac 1b – frac {ab}b=frac {1-ab}{b} right)$}\
&text{} && text{$left( frac 1a + frac 1b=frac b{ab} + frac a{ab}=frac {b+a}{ab} right)$}\
&= frac {b (b+a)}{ab (1-ab)} && text{$left(frac {frac ab}{frac cd}=frac {a cdot d}{b cdot c} right)$}\
&= frac {b+a}{a (1-ab)}
end{align*}
$$
$$
frac {frac 1a + frac 1b}{frac 1b – a} = frac {b+a}{a (1-ab)}
$$
b) $frac {frac 1a + frac 1b}{frac 1b – a} = frac {b+a}{a (1-ab)}$
* $bullet$ (0,-5)
* $bullet$ (4,3)
* $bullet$ (8,3)
$textbf{(b)}$ Below is rough sketch of the height of the rocket over time:
$$y=ax^2+bx+c$$
Using data points we have below three equations:
begin{itemize}
item $bullet$ $c=-5$
item $bullet$ $16a+4b+c=3$
item $bullet$ $64a+8b+c=3$
end{itemize}
Solving for $a$ and $b$:
begin{align*}
16a+4b+c&=64a+8b+c\
-48a&=4b\
b&=-12a
intertext{Putting $b=-12a$ and $c=-5$ in $16a+4b+c=3$:}
16a+4(-12a)-5&=3\
16a-48a&=8\
-32a&=8\
textcolor{blue}{a}&textcolor{blue}{=-dfrac{1}{4}}\
b&=-12a\
&=-12times left(-dfrac{1}{4} right)\
&=textcolor{blue}{3}
end{align*}
Hence equation of parabola is: $y=-frac{1}{4}x^{2}+3x-5$
Hence, rocket will hit the ground at $text{textcolor{#4257b2}{$t=10$ sec}}$
$textbf{(e)}$ Domain of the function is $color{#4257b2}{0le x le 10}$ as at $x=10$ rocket hits the ground and stops
$textbf{(f)}$ For $color{#4257b2}{0le x le 2}$, rocket is below ground
b.
$$
-5=c
$$
$$
3=16a+4b+c
$$
$$
3=64a+8b+c
$$
e. The domain from the function contains all possible $x$-values that make sense and thus is ${x|0leq xleq 10}$.
f. The function is negative on the domain at ${x|0leq xleq 2}$.
b. See graph
c. $f(x)=-dfrac{1}{4}x^2+3x-5$
d. 10 s
e. ${x|0leq xleq 10}$
f. ${x|0leq xleq 2}$
cos x=dfrac{1}{2}
$$
$x_2=dfrac{5pi}{3}$
The general solution is:
$x_1=dfrac{pi}{3}+2npi$
$x_2=dfrac{5pi}{3}+2npi, n$ integer
$x_2=dfrac{5pi}{3}+2npi, n$ integer
$x_2=dfrac{5pi}{3}, n$ integer
$x_2=dfrac{5pi}{3}+2npi, n$ integer
cos x=dfrac{1}{2}
$$
$x_2=dfrac{5pi}{3}$
$x_2=dfrac{5pi}{3}+2npi, n$ integer
Now if we put cosine inverse on both sides of the equation, we get,
$$
begin{align*}
cos ^{-1} (cos x) &= cos ^{-1} dfrac{1}{2} \\
x &= 60 text{textdegree} tag{$cos ^{-1} dfrac{1}{2}=60 text{textdegree}$} \
end{align*}
$$
Thus, x= $60 text{textdegree}$
In order to convert into radians, we can multiply the degree by $dfrac{pi}{180}$
Therefore in terms of radians x=$dfrac{pi}{180} times 60 text{textdegree}$ or $dfrac{pi}{3}$
This implies $cos(360-theta) = costheta$
Therefore,
$$
begin{align*}
cos 60 text{textdegree} &= cos(360-60 text{textdegree}) \
& = cos 300 text{textdegree} \
end{align*}
$$
This implies x = 300 $text{textdegree}$ as well.
In order to convert into radians, we can multiply the degree by $dfrac{pi}{180}$
Therefore in terms of radians x=$dfrac{pi}{180} times 300 text{textdegree}$ or $dfrac{5pi}{3}$
As we can observe that the value of $cos$ x for $cos x = dfrac{1}{2}$ repeats when we subtract $dfrac{pi}{3}$ and $dfrac{5pi}{3}$ from 2$pi$ or add 2$pi$ ‘n’ number of times to both of them respectively i.e the period of cosine is $2pi$
That is to say, $cos(2 pi(n) – theta) = cos theta$ and $cos(2 pi(n) + theta) = cos theta$
For example,
when n = 2
$cos (2pi(2) + dfrac{pi}{3} ) = cos (4 pi + dfrac{pi}{3}) = cos dfrac{13 pi}{3}$
and we know $cos(2 pi(n) + theta) = cos theta$
So, $cos dfrac{13 pi}{3} = cos dfrac{pi}{3} = dfrac{1}{2}$
Now we also have x=$dfrac{13pi}{3}$
x = $2pi(n) pm dfrac{pi}{3}$ and $2pi(n) pm dfrac{5pi}{3}$
Thus, if you put any value of n you will get $cos(x)= dfrac{1}{2}$
text{Given function in problem : }cos (x)=frac{1}{2}
$$
We know that the value of $cos (x)$ is $frac{1}{2}$ for two value of $x$ which are $frac{pi}{3}$ and $frac{5pi}{3}$ in interval $[0,pi]$ and period of the $cos$ function is $2pi$. So all the values of $x$ will be :
$$
x=frac{pi}{3}pm 2npi & x=frac{5pi}{3}pm 2npi
$$
{$text{color{#4257b2}Now by using above expression the value of $cos^{-1}(frac{1}{2})$ will be $frac{pi}{3}$ and $frac{5pi}{3}$. But we know that range of $cos^{-1}(x)$ function is $left[frac{-pi}{2},frac{pi}{2}right]$. So the there will be only one value for $cos^{-1}(frac{1}{2})$ and it is $frac{pi}{3}$}$
cos^{-1}(frac{1}{2})=frac{pi}{3}
$$
a)
$$
begin{align*}
2 cos x +1 &= 0\
2 cos x &= -1\
cos x &= – frac 12
end{align*}
$$
We use the inverse function
$$
begin{align*}
x &= arccos left( – frac 12 right)\
x &= frac {2 pi}{3} qquad x= frac {4 pi}{3}
end{align*}
$$
The period of the $cos x$ function is $2 pi$
$$
x = frac {2 pi}{3} + 2k pi qquad x= frac {4 pi}{3} + 2k pi
$$
for any integer $k$.
$$
begin{align*}
tan x &= 1
end{align*}
$$
We use the inverse function
$$
begin{align*}
x &= arctan 1\
x &= frac {pi}{4} qquad x = frac {5pi}{4}
end{align*}
$$
The period of the $tan x$ function is $pi$
$$
x = frac {pi}{4} + k pi qquad x = frac {5pi}{4} + k pi
$$
for any integer $k$.
b) $x = frac {pi}{4} + k pi qquad x = frac {5pi}{4} + k pi$
2cos x+1=0
$$
$$
cos x=-dfrac{1}{2}
$$
$x_1=dfrac{2pi}{3}$
$x_2=dfrac{4pi}{3}$
$x_2=dfrac{4pi}{3}+2npi, n$ integer
tan x=1
$$
x_1=dfrac{pi}{4}
$$
sin^{-1} x=37text{textdegree}
$$
We graph the solution $theta_1=37text{textdegree}$.
In order to determine another solution we draw a parallel to the $x$-axis, which intersects the circle in $B$:
theta_2=180text{textdegree}-angle BOY=180text{textdegree}-37text{textdegree}=143text{textdegree}
$$
$theta=143text{textdegree}+2npi, n$ integer
{color{#4257b2}text{a)}}
$$
Solution to this example is given below
$$
begin{align*}
2sin left(xright)-1+1&=0+1&&boxed{text{Add }1 text{ to both sides}}\
2sin left(xright)&=1&&boxed{text{Simplify}}\
frac{2sin left(xright)}{2}&=frac{1}{2}&&boxed{text{Divide both sides by } 2}\
sin left(xright)&=frac{1}{2}&&boxed{text{Simplify}}\
x=frac{pi }{6}+2pi n,:x&=frac{5pi }{6}+2pi n&&boxed{text{Simplify}}\\
&boxed{{color{#c34632}x=frac{pi }{6}+2pi n,:x=frac{5pi }{6}+2pi n} }&&boxed{text{Final solution}}\
end{align*}
$$
{color{#4257b2}text{b)}}
$$
Solution to this example is given below
$$
begin{align*}
frac{2cos left(xright)}{2}&=frac{-sqrt{3}}{2}&&boxed{text{Divide both sides by } 2}\
cos left(xright)&=-frac{sqrt{3}}{2}&&boxed{text{Simplify}}\
x&=frac{5pi }{6}+2pi n,:x=frac{7pi }{6}+2pi n&&boxed{text{Simplify}}\\
&boxed{{color{#c34632}x=frac{5pi }{6}+2pi n,:x=frac{7pi }{6}+2pi n} }&&boxed{text{Final solution}}\
end{align*}
$$
{color{#4257b2}text{c)}}
$$
Solution to this example is given below
$$
begin{align*}
frac{2sin left(xright)}{2}&=frac{sqrt{2}}{2}&&boxed{text{Divide both sides by } 2}\
sin left(xright)&=frac{sqrt{2}}{2}&&boxed{text{Simplify}}\
x=frac{pi }{4}+2pi n,:x&=frac{3pi }{4}+2pi n&&boxed{text{Simplify}}\\
&boxed{{color{#c34632}x=frac{pi }{4}+2pi n,:x=frac{3pi }{4}+2pi n} }&&boxed{text{Final solution}}\
end{align*}
$$
{color{#4257b2}text{d)}}
$$
Solution to this example is given below
$$
begin{align*}
x&=0+2pi n&&boxed{text{Simplify}}\
x&=2pi n&&boxed{text{Add}}\\
&boxed{{color{#c34632}x=2pi n} }&&boxed{text{Final solution}}\
end{align*}
$$
color{#4257b2} text{ a) }x=frac{pi }{6}+2pi n,:x=frac{5pi }{6}+2pi n
$$
$$
color{#4257b2} text{ b) } x=frac{5pi }{6}+2pi n,:x=frac{7pi }{6}+2pi n
$$
$$
color{#4257b2} text{ c) }x=frac{pi }{4}+2pi n,:x=frac{3pi }{4}+2pi n
$$
$$
color{#4257b2} text{ d) }x=2pi n
$$
$theta_2=128text{textdegree}$
$$
theta_3=308text{textdegree}
$$
costheta_1=costheta_2=cos theta_3
$$
cos x=y
$$
$cos 128text{textdegree}=y_2$
$cos 308text{textdegree}=y_3$
The three problems they have to work are:
As $cos 52text{textdegree}>0$ and $cos 128text{textdegree}<0$, the problem Tamara solves is different.
$$
y = sin x
$$
$$
y = |sin x|
$$
$$
y = sin x
$$
$$
y = |sin x|
$$
$$
g(x)=|sin x|
$$
g(x)=begin{cases}
-f(x), f(x)<0\
f(x), f(x)geq 0
end{cases}
$$
$$
begin{align*}
x + 2y &= 4\
2x – y &= -7\
x + y + z &= -4
end{align*}
$$
First equation
$$
begin{align*}
x + 2y &= 4\
x = 4 &- 2y
end{align*}
$$
Substitution in the second equation
$$
begin{align*}
2x – y &= -7\
2 ( 4-2y) -y &= -7\
8 – 4y – y &= -7\
8 – 5y &= -7\
– 5y &= -15
end{align*}
$$
Multiply this equation by $(-1)$
$$
begin{align*}
5y &= 15\
y &= frac {15}{5}\
y &= 3
end{align*}
$$
Now, we can substitute $y=3$ in the first equation
$$
begin{align*}
x &= 4 – 2 cdot 3\
x &= 4-6\
x &= -2
end{align*}
$$
And, we can substitute $x=-2$ and $y=3$ in the third equation
$$
begin{align*}
-2 + 3 + z &= -4\
1 + z &= -4\
z &= -5
end{align*}
$$
The solution of the system of equations is
$$
begin{align*}
x &= -2\
y &= 3\
z &= -5
end{align*}
$$
begin{align*}
x &= -2\
y &= 3\
z &= -5
end{align*}
$$
$$
begin{align*}
color{#c34632}{a cdot ln b = ln b^a} tag {$*$}\
color{#c34632}{ln (a cdot b) = ln a + ln b} tag {$**$}\
color{#c34632}{ln frac ab = ln a – ln b} tag{$***$}
end{align*}
$$
we obtain
$$
begin{align*}
3 ln x + 4 ln 2 – 2 ln y &overset{(*)}= ln x^3 + ln 2^4 – ln y^2\
&= ln x^3 + ln 16 – ln y^2\
&overset{(**)}= ln (x^3 16) – ln y^2\
&overset{(***)}= ln frac {16x^3}{y^2}
end{align*}
$$
$$
begin{align*}
color{#c34632}{a cdot ln b = ln b^a} tag {$*$}\
color{#c34632}{ln (a cdot b) = ln a + ln b} tag {$**$}
end{align*}
$$
Then
$$
begin{align*}
ln [(x-3)(3x+2)]^3 &overset{(*)}= 3 cdot ln [(x-3)(3x+2)]\
&overset{(**)}= 3 cdot [ ln (x-3)+ ln (3x+2)]\
&= 3 ln (x-3) + 3 ln (3x+2)
end{align*}
$$
b) $ln [(x-3)(3x+2)]^3=3 ln (x-3) + 3 ln (3x+2)$
3ln x+4ln 2-2ln y
$$
$$
=ln x^3+ln 16-ln y^2
$$
$log_a x^m=mlog_a x$.
ln x^3+ln 16-ln y^2=ln(16x^3)-ln y^2
$$
$log_a (xy)=log_a x+log_a y$.
ln(16x^3)-ln y^2=lndfrac{16x^3}{y^2}
$$
$log_a dfrac{x}{y}=log_a x-log_a y$.
ln[(x-3)(3x+2)]^3
$$
ln[(x-3)(3x+2)]^3=3ln [(x-3)(3x+2)]
$$
$log_a x^m=mlog_a x$.
3ln [(x-3)(3x+2)]=3ln (x-3)+3ln (3x+2)
$$
$log_a (xy)=log_a x+log_a y$.
b) $3ln (x-3)+3ln (3x+2)$
$$
color{#c34632}{a+ar+ar^2+ cdot cdot cdot }
$$
Therefore
$$
6+3 + frac 32+ cdot cdot cdot=6+frac 62+frac 6{2^2}+ cdot cdot cdot
$$
where
$$
a=6 qquad text{and} qquad r= frac 12
$$
According to the formula
$$
color{#c34632}{ S= frac {a}{1-r}}
$$
we obtain
$$
begin{align*}
S &= frac {6}{1 – frac 12}\
S &= frac {6}{frac 12}\
S &= 2 cdot 6\
S &= 12
end{align*}
$$
$$
sum_{k=1}^{infty} left( frac 13 right)^k = frac 13 + frac 19 + frac {1}{27} + cdot cdot cdot = frac 13 + frac 13 cdot frac 13 + frac 13 cdot frac 1{3^2} + cdot cdot cdot
$$
where
$$
a= frac 13 qquad text{and} qquad r= frac 13
$$
According to the formula
$$
color{#c34632}{ S= frac {a}{1-r}}
$$
we obtain
$$
begin{align*}
S &= frac {frac 13}{1 – frac 13}\
S &= frac {frac 13}{frac 23}\
S &= frac 12
end{align*}
$$
b) $S=frac 12$
6+3+dfrac{3}{2}+….
$$
$a_n=dfrac{1}{2}a_{n-1}$
$$
q=dfrac{1}{2}
$$
S=dfrac{a_1}{1-q}=dfrac{6}{1-dfrac{1}{2}}=dfrac{6}{dfrac{1}{2}}=12
$$
sum_{k=1}^{infty}left(dfrac{1}{3}right)^k
$$
$a_n=dfrac{1}{3}a_{n-1}$
$$
q=dfrac{1}{3}
$$
S=dfrac{a_1}{1-q}=dfrac{dfrac{1}{3}}{1-dfrac{1}{3}}=dfrac{dfrac{1}{3}}{dfrac{2}{3}}=dfrac{1}{2}
$$
b) $dfrac{1}{2}$
begin{tabular}{|| c| c||}
hline
$x$ & $y=sin x$ \ [0.5ex]
hline
0 & 0 \
hline
$dfrac{pi}{4}$ & $dfrac{sqrt 2}{2}$ \
hline
$dfrac{pi}{2}$ & 1 \
hline
$dfrac{3pi}{4}$ & $dfrac{sqrt 2}{2}$ \
hline
$pi$ & 0 \
hline
$dfrac{5pi}{4}$ & $-dfrac{sqrt 2}{2}$ \
hline
$dfrac{3pi}{2}$ & -1 \
hline
$dfrac{7pi}{4}$ & $-dfrac{sqrt 2}{2}$ \
hline
$2pi$ & 0 \[1ex]
hline
end{tabular}
end{center}
Range: $[-1,1]$
We determine the domain and range:
Range: $[-1,1]$
Range:$left[-dfrac{pi}{2},dfrac{pi}{2}right]$
sin x=-dfrac{sqrt 3}{2}
$$
$x_2=dfrac{5pi}{3}$
$x_2=5.2359878=dfrac{5pi}{3}$
Range: $[-1,1]$
Range: $mathbb{R}$
Range: $[0,pi]$
Range: $left(-dfrac{pi}{2},dfrac{pi}{2}right)$
{color{#4257b2}text{a)}}
$$
Solution to this example is given below
$$
begin{align*}
2cos left(xright)-1+1&=0+1&&boxed{text{Add }1 text{ to both sides}}\
2cos left(xright)&=1&&boxed{text{Simplify}}\
frac{2cos left(xright)}{2}&=frac{1}{2}&&boxed{text{Divide both sides by } 2}\
cos left(xright)&=frac{1}{2}&&boxed{text{Simplify}}\
x=frac{pi }{3}+2pi n,:x&=frac{5pi }{3}+2pi n&&boxed{text{Simplify}}\\
&boxed{{color{#c34632}x=frac{pi }{3}+2pi n,:x=frac{5pi }{3}+2pi n} }&&boxed{text{Final solution}}\\
end{align*}
$$
Graph of $boldsymbol{2 cos(x)-1=0}$
{color{#4257b2}text{b)}}
$$
Solution to this example is given below
$$
begin{align*}
x&=frac{pi }{3}+pi n&&boxed{text{Simplify}}\\
&boxed{{color{#c34632}x=frac{pi }{3}+pi n} }&&boxed{text{Final solution}}\\
end{align*}
$$
Graph of $boldsymbol{tan(x)=sqrt3}$
{color{#4257b2}text{c)}}
$$
Solution to this example is given below
$$
begin{align*}
frac{2sin left(xright)}{2}&=frac{sqrt{3}}{2}&&boxed{text{Divide both sides by } 2}\
sin left(xright)&=frac{sqrt{3}}{2}&&boxed{text{Simplify}}\
x=frac{pi }{3}+2pi n,:x&=frac{2pi }{3}+2pi n&&boxed{text{Simplify}}\\
&boxed{{color{#c34632}x=frac{pi }{3}+2pi n,:x=frac{2pi }{3}+2pi n} }&&boxed{text{Final solution}}\\
end{align*}
$$
Graph of $boldsymbol{2 sin(x)=sqrt3}$
{color{#4257b2}text{d)}}
$$
Solution to this example is given below
$$
begin{align*}
4u^2-3&=0&&boxed{text{Let: }sin(x)=u }\
4u^2-3+3&=0+3&&boxed{text{Add 3 to both sides}}\
4u^2&=3&&boxed{text{Simplify}}\
frac{4u^2}{4}&=frac{3}{4}&&boxed{text{Divide both sides by } 4}\
u^2&=frac{3}{4}&&boxed{text{Simplify}}\
u=sqrt{frac{3}{4}},:u&=-sqrt{frac{3}{4}}&&boxed{text{Simplify}}\
u=frac{sqrt{3}}{2},:u&=-frac{sqrt{3}}{2}&&boxed{text{Simplify}}\
sin left(xright)&=frac{sqrt{3}}{2},:sin left(xright)=-frac{sqrt{3}}{2}&&boxed{text{Substitute back: } u=sin(x)}\
x=frac{pi }{3}+2pi n,:x=frac{2pi }{3}+2pi n,:x&=frac{4pi }{3}+2pi n,:x=frac{5pi }{3}+2pi n&&boxed{text{Combine the solutions}}\\
&boxed{{color{#c34632}x=frac{pi }{3}+2pi n,:x=frac{2pi }{3}+2pi n,:} }&&boxed{text{Final solution}}\
&boxed{{color{#c34632}x=frac{4pi }{3}+2pi n,:x=frac{5pi }{3}+2pi n} }&&boxed{text{Final solution}}\\
end{align*}
$$
Graph of $boldsymbol{4sin^2(x)-3=0}$
color{#4257b2} text{ a) } x=frac{pi }{3}+2pi n,:x=frac{5pi }{3}+2pi n
$$
$$
color{#4257b2} text{ b) } x=frac{pi }{3}+pi n
$$
$$
color{#4257b2} text{ c) } x=frac{pi }{3}+2pi n,:x=frac{2pi }{3}+2pi n
$$
$$
color{#4257b2} text{ d) }x=frac{pi }{3}+2pi n,:x=frac{2pi }{3}+2pi n,:
$$
$$
color{#4257b2} text{ d) } x=frac{4pi }{3}+2pi n,:x=frac{5pi }{3}+2pi n
$$
$$
sin theta = 0.5
$$
By applying the inverse function we get
$$
begin{align*}
theta &= arcsin 0.5\
theta &= frac {pi}{6} qquad theta = frac {5 pi}{6}
end{align*}
$$
The period of the $sin x$ function is $2pi$
$$
theta = frac {pi}{6} + 2k pi qquad theta = frac {5 pi}{6} + 2k pi
$$
for any integer $k$.
$$
cos theta = – 0.5
$$
By applying the inverse function we get
$$
begin{align*}
theta &= arccos (-0.5)\
theta &= frac {2pi}{3} qquad theta = frac {4 pi}{3}
end{align*}
$$
The period of the $cos x$ function is $2pi$
$$
theta = frac {2pi}{3} + 2k pi qquad theta = frac {4 pi}{3} + 2k pi
$$
$$
begin{align*}
4 tan theta – 4 &= 0\
4 tan theta &= 4\
tan theta &= 1
end{align*}
$$
By applying the inverse function we get
$$
begin{align*}
theta &= arctan 1\
theta &= frac {pi}{4} qquad theta = frac {5 pi}{4}
end{align*}
$$
The period of the $tan x$ function is $pi$
$$
theta = frac {pi}{4} + k pi qquad theta = frac {5 pi}{4} + k pi
$$
b) $theta = frac {2pi}{3} + 2k pi qquad theta = frac {4 pi}{3} + 2k pi$
c) $theta = frac {pi}{4} + k pi qquad theta = frac {5 pi}{4} + k pi$
$$
y=cos (3x)-1
$$
is the same as a graph under $ii$.
$$
y=3cos (x)-1
$$
is the same as a graph under $iii$.
$$
y=3cos (x-1)
$$
is the same as a graph under $iv$.
$$
y=3cos (3x-1)
$$
is the same as a graph under $i$.
b) $iii$
c) $iv$
d) $i$
$$
begin{align*}
frac {3x}{x^2 + 2x +1} + frac {3}{x^2 + 2x +1} &= frac {3x + 3}{x^2 + 2x +1}\
&overset{(*)}= frac {3(x+1)}{(x+1)^2}\
&= frac {3}{x+1}
end{align*}
$$
Using the formula
$$
{(star)} x^2 + 2x + 1 = (x+1)^2
$$
Therefore
$$
frac {3x}{x^2 +2x+1} + frac {3}{x^2 + 2x+1} = frac {3}{x+1}
$$
$$
begin{align*}
frac {3}{x-1} – frac {2}{x-2} &= frac {3(x-2)-2(x-1)}{(x-1)(x-2)}\
&= frac {3x-6-2x+2}{x^2 -2x-x+2}\
&= frac {x-4}{x^2 -3x +2}
end{align*}
$$
Therefore
$$
frac {3}{x-1} – frac {2}{x-2} = frac {x-4}{x^2 -3x +2}
$$
b) $frac {3}{x-1} – frac {2}{x-2} = frac {x-4}{x^2 -3x +2}$
begin{tabular}{|| c| c||}
hline
x & f(x) \ [0.5ex]
hline
-2 & 3 \
hline
-1 & 4\
hline
0 & 5\
hline
1 & undefined\
hline
2 & 7\
hline
3 & 8\[1ex]
hline
end{tabular}
end{center}
It is not correct to connect all the dots because the function is not defined for $x=1$.
$f(1.1)=dfrac{1.1^2+4(1.1)-5}{1.1-1}=6.1$
$f(x)=x+5$.
We compute $f(0.9)$ and $f(1.1)$:
$f(1.01)=dfrac{1.01^2+4(1.01)-5}{1.01-1}=6.01$
$=dfrac{x(x-1)+5(x-1)}{x-1}$
$=dfrac{(cancel{x-1})(x+5)}{cancel{x-1}}$
$$
=x+5
$$
$y=ax^2+bx+c$
$24=a(-2)^2+b(-2)+c$ (Substituting with the point $(-2, 24)$)
$24=4a-2b+c$ (1)
$-1=a(3)^2+b(3)+c$ (Substituting with the point $(3, -1)$)
$-1=9a+3b+c$ (2)
$15=a(-1)^2+b(-1)+c$ (Substituting with the point $(-1, 15)$)
$15=a-b+c$ (3)
The system is:
$24=4a-2b+c$ (1)
$-1=9a+3b+c$ (2)
$15=a-b+c$ (3)
Subtracting (1) from (2)
$-1=9a+3b+c$ (2)
–
$24=4a-2b+c$ (1)
—————————-
$-25=5a+5b$
$-5=a+b$ (4)
Subtracting (2) from (3)
$15=a-b+c$ (3)
–
$-1=9a+3b+c$ (2)
—————————-
$16=-8a-4b$
$4=-2a-b$ (5)
$-5=a+b$ (4)
+
$4=-2a-b$ (5)
————————–
$-1=-a$
$$
a=1
$$
Substituting in (4)
$-5=1+b$
$b=-5-12$
$$
b=-6
$$
Substituting in (1)
$24=4a-2b+c$ (1)
$24=4(1)-2(-6)+c$
$24=4+12+c$
$c=24-16$
$$
c=8
$$
The solution is: $a=1$ $b=-6$ $c=8$
The equation is:
$$
y=x^2-6x+8
$$
$24=4a-2b+c$ (1)
$24=4+12+8$ checkmark
$-1=9a+3b+c$ (2)
$-1=9(1)+3(-6)+8$
$-1=9-18+8$
$-1=17-18$ checkmark
$15=a-b+c$ (3)
$15=1-(-6)+8$
$15=7+8$ checkmark
$$
begin{align*}
frac {frac {x+1}{2x}}{frac {x^2 -1}{x}} &= frac {x(x+1)}{2x(x^2 -1)}\
&overset{(*)}= frac {x+1}{2(x-1)(x+1)}\
&= frac {1}{2(x-1)}
end{align*}
$$
Using the formula
$$
{(star)} x^2 -1=(x-1)(x+1)
$$
Therefore
$$
frac {frac {x+1}{2x}}{frac {x^2 -1}{x}} = frac {1}{2(x-1)}
$$
$$
begin{align*}
frac {frac {4}{x+3}}{frac 1x +3} = frac {frac {4}{x+3}}{frac {1+3x}{x}} &= frac {4x}{(x+3)(1+3x)}\
&= frac {4x}{x+3x^2 + 3 +9x}\
&= frac {4x}{3x^2 + 10x +3}
end{align*}
$$
Therefore
$$
frac {frac {4}{x+3}}{frac 1x +3} = frac {4x}{3x^2 + 10x +3}
$$
b) $frac {frac {4}{x+3}}{frac 1x +3} = frac {4x}{3x^2 + 10x +3}$
$$
sum_{n=1}^8 3^n= 3+3 cdot 3 + 3 cdot 3^2 + 3 cdot 3^3 + cdot cdot cdot + 3 cdot 3^7
$$
where
$$
a=3 qquad text{and} qquad r=3 qquad text{and} qquad k=7
$$
According to the formula
$$
color{#c34632}{ S = a left( frac {1- r^{k+1}}{1-r} right) }
$$
we obtain
$$
begin{align*}
S &= 3 left( frac {1- 3^{7+1}}{1-3} right)\
S &= 3 frac {1- 3^8}{-2}\
S &= 3 frac {1-6561}{-2}\
S &= 3 frac {-6560}{-2}\
S &= 3 cdot 3280\
S &= 9840
end{align*}
$$
The sum is
$$
sum_{n=1}^8 3^n=9840
$$
$$
begin{align*}
sum_{c=1}^{16} (3c + 5) &= (3 cdot 1 +5) + (3 cdot 2 + 5) +cdot cdot cdot+ (3 cdot 16 +5)\
&= (3+5) + (6+5) + cdot cdot cdot+ (48 +5)\
&= 8+ 11 +cdot cdot cdot+ 53
end{align*}
$$
where
$$
a=8 qquad text{and} qquad d=11-8=3 qquad text{and} qquad n=16
$$
According to the formula
$$
color{#c34632}{ S= frac n2 [ 2a + (n-1) d]}
$$
we obtain
$$
begin{align*}
S &= frac {16}{2} [ 2 cdot 8 + (16-1) cdot 3]\
S &= 8 [ 16 + 15 cdot 3]\
S &= 8 [ 16 + 45 ]\
S &= 8 cdot 61\
S &= 488
end{align*}
$$
Therefore
$$
sum_{c=1}^{16} (3c + 5) = 488
$$
b) $sum_{c=1}^{16} (3c + 5) = 488$
$$y = sin x$$
and it’s inverse
$$y = sin^{-1} x$$
$rightarrow y = sin x$ is
$$text {Domain }: left[-dfrac{pi}{2}, dfrac{pi}{2} right]$$
$$text {Range }: [-1,1]$$
$$text {Domain }: [-1,1]$$
$$text {Range }: left[-dfrac{pi}{2}, dfrac{pi}{2} right]$$
$$y = cos x$$
and it’s inverse
$$y = cos^{-1} x$$
$rightarrow y = cos x$ is
$$text {Domain }: [0, pi]$$
$$text {Range }: [-1,1]$$
$$text {Domain }: [0, pi]$$
$$text {Range }: [-1,1]$$
$$y = tan x$$
and it’s inverse
$$y = tan^{-1} x$$
$rightarrow y = tan x$ is
$$text {Domain }: left(-dfrac{pi}{2}, dfrac{pi}{2} right)$$
$$text {Range }: Reals$$
$$text {Domain }: Reals$$
$$text {Range }: left(-dfrac{pi}{2}, dfrac{pi}{2} right)$$
{color{#4257b2}text{a)}}
$$
Solution to this example is given below
$$
begin{align*}
4sin left(xright)+2-2&=0-2&&boxed{text{Subtract }2 text{ from both sides}}\
4sin left(xright)&=-2&&boxed{text{Simplify}}\
frac{4sin left(xright)}{4}&=frac{-2}{4}&&boxed{text{Divide both sides by } 4}\
sin( x) &=-frac{2}{4}&&boxed{text{Simplify}}\
sin left(xright)&=-frac{1}{2}&&boxed{text{Cross cancel common factor: }2}\
x=frac{7pi }{6}+2pi n,:x&=frac{11pi }{6}+2pi n&&boxed{text{Simplify}}\\
&boxed{{color{#c34632}x=frac{7pi }{6}+2pi n,:x=frac{11pi }{6}+2pi n} }&&boxed{text{Final solution}}\
end{align*}
$$
Graph of $boldsymbol{4 sin (x)+2=0}$
{color{#4257b2}text{b)}}
$$
Solution to this example is given below
$$
begin{align*}
frac{2cos left(xright)}{2}&=frac{sqrt{3}}{2}&&boxed{text{Divide both sides by } 2}\
cos left(xright)&=frac{sqrt{3}}{2}&&boxed{text{Simplify}}\
x=frac{pi }{6}+2pi n,:x&=frac{11pi }{6}+2pi n&&boxed{text{Simplify}}\\
&boxed{{color{#c34632}x=frac{pi }{6}+2pi n,:x=frac{11pi }{6}+2pi n} }&&boxed{text{Final solution}}\
end{align*}
$$
Graph of $boldsymbol{2 cos (x)=sqrt3}$
{color{#4257b2}text{c)}}
$$
Solution to this example is given below
$$
begin{align*}
tan left(xright)+1-1&=0-1&&boxed{text{Subtract 1 from both sides}}\
tan left(xright)&=-1&&boxed{text{Simplify}}\
x&=frac{3pi }{4}+pi n&&boxed{text{Simplify}}\\
&boxed{{color{#c34632}x=frac{3pi }{4}+pi n} }&&boxed{text{Final solution}}\
end{align*}
$$
Graph of $boldsymbol{tan(x)+1=0}$
{color{#4257b2}text{d)}}
$$
Solution to this example is given below
$$
begin{align*}
4u^2-4&=0&&boxed{text{Let: }cos(x)=u }\
4u^2-4+4&=0+4&&boxed{text{Add 4 to both sides}}\
4u^2&=4&&boxed{text{Simplify}}\
frac{4u^2}{4}&=frac{4}{4}&&boxed{text{Divide both sides by 4}}\
u^2&=1&&boxed{text{Simplify}}\
u=sqrt{1},:u&=-sqrt{1}&&boxed{text{Simplify}}\
u=1,:u&=-1&&boxed{text{Simplify}}\
cos left(xright)&=1,:cos left(xright)=-1&&boxed{text{Substitute back: }u=cos(x)}\
end{align*}
$$
First we solve $cos(x)=1$
$$
begin{align*}
x&=0+2pi n&&boxed{text{Simplify}}\
x&=2pi n&&boxed{text{Add }}\
end{align*}
$$
Second we solve $cos(x)=-1$
$$
begin{align*}
x&=pi +2pi n&&boxed{text{Simplify}}\\
&boxed{{color{#c34632}x=2pi n,:x=pi +2pi n} }&&boxed{text{Combine the solutions}}\
end{align*}
$$
Graph of $boldsymbol{4 cos^2(x)-4=0}$
color{#4257b2} text{ a) } x=frac{7pi }{6}+2pi n,:x=frac{11pi }{6}+2pi n
$$
$$
color{#4257b2} text{ b) } x=frac{pi }{6}+2pi n,:x=frac{11pi }{6}+2pi n
$$
$$
color{#4257b2} text{ c) }x=frac{3pi }{4}+pi n
$$
$$
color{#4257b2} text{ d) } x=2pi n,:x=pi +2pi n
$$
f(theta)=tantheta
$$
g(theta)=tantheta+1
$$
h(theta)=tanleft(theta+dfrac{pi}{4}right)
$$
j(theta)=-tantheta
$$
k(theta)=4tantheta
$$
{color{#4257b2}text{a)}}
$$
Solution to this example is given below
$$
begin{align*}
&frac{1cdot3}{6}+frac{1cdot2}{6}&&boxed{text{Take LCM for 2 and 3}}\
&frac{3}{6}+frac{2}{6}&&boxed{text{Simplify}}\
&frac{3+2}{6}&&boxed{text{Combine the numerators}}\
&frac{5}{6}&&boxed{text{Simplify}}\\
&boxed{{color{#c34632}frac{5}{6}} }&&boxed{text{Final solution}}\
end{align*}
$$
$$
boxed{ color{#c34632} text{ } mathrm{Since:the:denominators:are:equal,:combine:the:numerators}:quad frac{a}{c}pm frac{b}{c}=frac{apm :b}{c}}
$$
{color{#4257b2}text{b)}}
$$
Solution to this example is given below
$$
begin{align*}
&frac{3x}{2xx}+frac{4}{x^2}&&boxed{text{For } frac{3}{2x }text{ Multiply the numerator and denominator by }x}\
&frac{3x}{2x^2}+frac{4}{x^2}&&boxed{text{Simplify}}\
&frac{3x}{2x^2}+frac{4cdot :2}{x^2cdot :2}&&boxed{text{For } frac{4}{x^2}: text{ Multiply the numerator and denominator by 2}}\
&frac{3x}{2x^2}+frac{8}{2x^2}&&boxed{text{Simplify}}\
&frac{3x+8}{2x^2}&&boxed{text{Combine the numerators}}\\
&boxed{{color{#c34632}frac{3x+8}{2x^2}} }&&boxed{text{Final solution}}\
end{align*}
$$
$$
boxed{ color{#c34632} text{ } mathrm{Since:the:denominators:are:equal,:combine:the:numerators}:quad frac{a}{c}pm frac{b}{c}=frac{apm :b}{c}}
$$
{color{#4257b2}text{c)}}
$$
Solution to this example is given below
$$
begin{align*}
&frac{xleft(x-1right)}{left(x+1right)left(x-1right)}+frac{3}{x-1}
&&boxed{text{For } frac{x}{x+1 }text{ Multiply the numerator and denominator by }x-1}\
&frac{xleft(x-1right)}{left(x+1right)left(x-1right)}+frac{3left(x+1right)}{left(x-1right)left(x+1right)}&&boxed{text{For } frac{3}{x-1}: text{ Multiply the numerator and denominator by }x+1}\
&frac{xleft(x-1right)+3left(x+1right)}{left(x+1right)left(x-1right)}&&boxed{text{Combine the numerators}}\
&frac{x^2-x+3x+3}{left(x+1right)left(x-1right)}&&boxed{text{Distributive property}}\
&frac{x^2+2x+3}{left(x+1right)left(x-1right)}&&boxed{text{Combine like terms: } -x+3x=2x}\\
&boxed{{color{#c34632}frac{x^2+2x+3}{left(x+1right)left(x-1right)}} }&&boxed{text{Final solution}}\
end{align*}
$$
$$
boxed{ color{#c34632} text{ } mathrm{Since:the:denominators:are:equal,:combine:the:numerators}:quad frac{a}{c}pm frac{b}{c}=frac{apm :b}{c}}
$$
{color{#4257b2}text{d)}}
$$
Solution to this example is given below
$$
begin{align*}
&frac{sin ^2left(thetaright)}{cos left(thetaright)sin left(thetaright)}+frac{cos left(thetaright)}{cos left(thetaright)sin left(thetaright)}&&boxed{text{Adjust fraction based on the LCM}}\
&frac{sin ^2left(thetaright)+cos left(thetaright)}{cos left(thetaright)sin left(thetaright)}&&boxed{text{Add the fraction}}\
&frac{cos left(thetaright)+sin ^2left(thetaright)}{frac{sin left(2thetaright)}{2}}&&boxed{text{Use the following identity}}\
&frac{2left(sin ^2left(thetaright)+cos left(thetaright)right)}{sin left(2thetaright)}&&boxed{text{Simplify}}\\
&boxed{{color{#c34632}frac{2left(sin ^2left(thetaright)+cos left(thetaright)right)}{sin left(2thetaright)}} }&&boxed{text{Final solution}}\
end{align*}
$$
$$
boxed{ color{#c34632} text{ } mathrm{Use:the:following:identity}:quad cos left(xright)sin left(xright)=frac{sin left(2xright)}{2}}
$$
$$
boxed{ color{#c34632} text{ }mathrm{Apply:the:fraction:rule}:quad frac{a}{frac{b}{c}}=frac{acdot :c}{b} }
$$
color{#4257b2} text{ a) } frac{5}{6}
$$
$$
color{#4257b2} text{ b) }frac{3x+8}{2x^2}
$$
$$
color{#4257b2} text{ c) } frac{x^2+2x+3}{left(x+1right)left(x-1right)}
$$
$$
color{#4257b2} text{ d) } frac{2left(sin ^2left(thetaright)+cos left(thetaright)right)}{sin left(2thetaright)}
$$
$$
(x-a)^2+(y-b)^2=r^2
$$
with $(a,b)$ the center of the circle and $r$ the radius of the circle.
a. Replace $r$ with $2$ and $b$ with $3a$ (because the cetner has to lie on the line $y=3x$):
$$
(x-a)^2+(y-3a)^2=4
$$
The function has to be tangent to the $y$-axis and thus $a=2$:
$$
(x-2)^2+(y-6)^2=4
$$
b. Replace $r$ with $3$ and $b$ with $3a$ (because the cetner has to lie on the line $y=3x$):
$$
(x-a)^2+(y-3a)^2=9
$$
The function has to be tangent to the $y$-axis and thus $a=3$:
$$
(x-3)^2+(y-9)^2=9
$$
b. $(x-3)^2+(y-9)^2=9$
y=-3-x
$$
2x-(-3-x)=-6
$$
3(-3)-2(0)+5z=16
$$
z=5
$$
$$
color{#c34632}{c^2 = a^2 + b^2 – 2ab cos theta}
$$
Then
$$
begin{align*}
c^2 &= 412^2 + 348^2 – 2 cdot 412 cdot 348 cdot cos 39^{text{textdegree}}\
c^2 &= 169 744 + 121 104 – 286752 cdot 0.77\
c^2 &= 290 848 – 220 799.04\
c^2 &= 70 048.96\
c &= 264.66
end{align*}
$$
c = 264.66
$$
$BL^2=412^2+348^2-2cdot 412cdot 348cdot 0.77714596$
$BL^2=67999.842$
$BL=sqrt{67999.842}$
$BLapprox 260.77$ feet
We use the Law of Cosines to determine $angle B$:
$$
g(x)=dfrac{1}{sin x}
$$
begin{tabular}{|| c| c||}
hline
$x$ & $f(x)=sin x$ \ [0.5ex]
hline
0 & 0 \
hline
$dfrac{pi}{4}$ &0.70 \
hline
$dfrac{pi}{2}$ & 1 \
hline
$dfrac{3pi}{4}$ & 0.70 \
hline
$pi$ & 1 \
hline
$dfrac{5pi}{4}$ & -0.70 \
hline
$dfrac{3pi}{2}$ & -1 \
hline
$dfrac{7pi}{4}$ & -0.70 \
hline
$2pi$ & 0 \[1ex]
hline
end{tabular}
end{center}
begin{tabular}{|| c|c| c||}
hline
$x$ & $f(x)=sin x$ & $g(x)=dfrac{1}{sin x}$ \ [0.5ex]
hline
0 & 0 & undefined \
hline
$dfrac{pi}{4}$ &0.70 & 1.41\
hline
$dfrac{pi}{2}$ & 1 & 1 \
hline
$dfrac{3pi}{4}$ & 0.70& 1.41 \
hline
$pi$ & 0& undefined \
hline
$dfrac{5pi}{4}$ & -0.70& -1.41 \
hline
$dfrac{3pi}{2}$ & -1 &-1\
hline
$dfrac{7pi}{4}$ & -0.70& -1.41 \
hline
$2pi$ & 0 & undefined\[1ex]
hline
end{tabular}
end{center}
The reciprocal values of $sin 0,sin pi$ do not make sense,so the graph of $g(x)$ will have vertical asymptotes in $x=0, x=pi$, while the reciprocal value of $sin dfrac{pi}{2}$ will be a local minimum of $g(x)$.
$csc x=dfrac{1}{sin x}$.
(-infty,-1]cup[1,infty)
$$
mathbb{R}-{kpi|ktext{ integer}}
$$
Range of $y=sin x$: $[-1,1]$
$x_2=dfrac{5pi}{6}$
$x_2=dfrac{5pi}{6}$
x=dfrac{pi}{2}
$$
x=dfrac{pi}{2}
$$
$x_2=dfrac{2pi}{3}$
$x_2=dfrac{2pi}{3}$
$$
csc x=dfrac{1}{a}
$$
csc x=dfrac{1}{sin x}
$$
We have to find $csc (x)=3.5$
We know, $csc theta = dfrac{1}{sin theta}$
$$
begin{align*}
csc (x)&=3.5 \
dfrac{1}{sin (x)} &=3.5 \\
sin(x)&=dfrac{1}{3.5} \\
sin ^{-1} [(sin (x) ]&=sin ^{-1} left(dfrac{10}{35}right) tag {Taking $sin ^{-1}$ on both sides }\\
x&=sin ^{-1} left(dfrac{2}{7}right) \
end{align*}
$$
Since $sin ^{-1}$ is given in the calculator we can find its value.
Thus, $x=sin ^{-1} left(dfrac{2}{7}right) = 0.29$
Therefore, for $csc (x)=3.5$, x = 0.29.
$$
g(x)=dfrac{1}{cos x}
$$
begin{tabular}{|| c| c||}
hline
$x$ & $f(x)=cos x$ \ [0.5ex]
hline
0 & 1 \
hline
$dfrac{pi}{4}$ &0.70 \
hline
$dfrac{pi}{2}$ & 0 \
hline
$dfrac{3pi}{4}$ & -0.70 \
hline
$pi$ & 0 \
hline
$dfrac{5pi}{4}$ & -0.70 \
hline
$dfrac{3pi}{2}$ & 0 \
hline
$dfrac{7pi}{4}$ & 0.70 \
hline
$2pi$ & 1 \[1ex]
hline
end{tabular}
end{center}
begin{tabular}{|| c|c| c||}
hline
$x$ & $f(x)=cos x$ & $g(x)=dfrac{1}{cos x}$ \ [0.5ex]
hline
0 & 1 & 1 \
hline
$dfrac{pi}{4}$ &0.70 & 1.41\
hline
$dfrac{pi}{2}$ & 0 & undefined \
hline
$dfrac{3pi}{4}$ & -0.70& -1.41 \
hline
$pi$ & -1& -1 \
hline
$dfrac{5pi}{4}$ & -0.70& -1.41 \
hline
$dfrac{3pi}{2}$ & 0 &undefined\
hline
$dfrac{7pi}{4}$ & 0.70& 1.41 \
hline
$2pi$ & 1 & 1\[1ex]
hline
end{tabular}
end{center}
$$
sec x=dfrac{1}{cos x}
$$
Range of $y=sec x$: $(-infty,-1]cup[1,infty)$
$$
g(x)=dfrac{1}{tan x}
$$
begin{tabular}{|| c|c| c||}
hline
$x$ & $f(x)=tan x$ & $g(x)=dfrac{1}{tan x}$ \ [0.5ex]
hline
0 & 0 & undefined \
hline
$dfrac{pi}{6}$ &0.58 & 1.731\
hline
$dfrac{pi}{3}$ & 1.73 & 0.58 \
hline
$dfrac{pi}{2}$ & undefined& 0 \
hline
$dfrac{2pi}{3}$ & -1.73& -0.58 \
hline
$dfrac{3pi}{4}$ & -1& -1 \
hline
$pi$ & 0 & undefined\[1ex]
hline
end{tabular}
end{center}
$$
cot x=dfrac{1}{tan x}
$$
Range: $mathbb{R}$
{color{#4257b2}text{a)}}
$$
Solution to this example is given below
$$
begin{align*}
sin ^2left(xright)csc left(xright)-frac{sqrt{3}}{2}&=0&&boxed{text{Subtract }frac{sqrt{3}}{2}text{ from both sides}}\
frac{sin ^2left(xright)csc left(xright)cdot :2}{2}-frac{sqrt{3}}{2}&=0&&boxed{text{Convert element to a fraction}}\
frac{sin ^2left(xright)csc left(xright)cdot :2-sqrt{3}}{2}&=0&&boxed{text{Combine the numerators}}\
frac{2sin ^2left(xright)csc left(xright)-sqrt{3}}{2}&=0&&boxed{text{Simplify}}\
2sin ^2left(xright)csc left(xright)-sqrt{3}&=0&&boxed{text{Apply rule}}\
-sqrt{3}+2cdot frac{1}{sin left(xright)}sin ^2left(xright)&=0&&boxed{text{Use the following identity}}\
-sqrt{3}+2sin left(xright)&=0&&boxed{text{Simplify}}\
-sqrt{3}+2sin left(xright)+sqrt{3}&=0+sqrt{3}&&boxed{text{Add } sqrt3 text{to both sides}}\
2sin left(xright)&=sqrt{3}&&boxed{text{Simplify}}\
frac{2sin left(xright)}{2}&=frac{sqrt{3}}{2}&&boxed{text{Divide both sides }3}\
sin left(xright)&=frac{sqrt{3}}{2}&&boxed{text{Simplify}}\
x=frac{pi }{3}+2pi n,&:x=frac{2pi }{3}+2pi n&&boxed{text{Simplify}}\
end{align*}
$$
The equation is $textbf{ Sometimes true}$
$$
begin{align*}
&boxed{{color{#c34632}x=frac{pi }{3}+2pi n,:x=frac{2pi }{3}+2pi n} }&&boxed{text{Final solution}}\
end{align*}
$$
$$
boxed{ color{#c34632} text{ }frac{fleft(xright)}{gleft(xright)}=0quad Rightarrow quad fleft(xright)=0 }
$$
$$
boxed{ color{#c34632} text{ } mathrm{Since:the:denominators:are:equal,:combine:the:numerators}:quad frac{a}{c}pm frac{b}{c}=frac{apm :b}{c}}
$$
{color{#4257b2}text{b)}}
$$
Solution to this example is given below
$$
begin{align*}
sec ^2left(xright)-frac{2}{cos left(xright)}&=0&&boxed{text{Subtract }frac{2}{cos(x)}text{ from both sides}}\
frac{sec ^2left(xright)cos left(xright)}{cos left(xright)}-frac{2}{cos left(xright)}&=0&&boxed{text{Convert element to a fraction}}\
frac{sec ^2left(xright)cos left(xright)-2}{cos left(xright)}=0&=0&&boxed{text{Combine the numerators}}\
sec ^2left(xright)cos left(xright)-2&=0&&boxed{text{Apply rule}}\
-2+left(frac{1}{cos left(xright)}right)^2cos left(xright)&=0&&boxed{text{Use the following identity}}\
-2+frac{1}{cos left(xright)}&=0&&boxed{text{Simplify}}\
-2cos left(xright)+frac{1}{cos left(xright)}cos left(xright)&=0cdot cos left(xright)&&boxed{text{Multiply both sides by }cos(x)}\
-2cos left(xright)+1&=0&&boxed{text{Simplify}}\
-2cos left(xright)+1-1&=0-1&&boxed{text{Subtract 1 from both sides}}\
-2cos left(xright)&=-1&&boxed{text{Simplify}}\
frac{-2cos left(xright)}{-2}&=frac{-1}{-2}&&boxed{text{Divide both sides }-1}\
cos left(xright)&=frac{1}{2}&&boxed{text{Simplify}}\
x=frac{pi }{3}+2pi n,&:x=frac{5pi }{3}+2pi n&&boxed{text{Simplify}}\
end{align*}
$$
The equation is $textbf{ Sometimes true}$
$$
begin{align*}
&boxed{{color{#c34632}x=frac{pi }{3}+2pi n,:x=frac{5pi }{3}+2pi n} }&&boxed{text{Final solution}}\
end{align*}
$$
$$
boxed{ color{#c34632} text{ }frac{fleft(xright)}{gleft(xright)}=0quad Rightarrow quad fleft(xright)=0 }
$$
$$
boxed{ color{#c34632} text{ } mathrm{Since:the:denominators:are:equal,:combine:the:numerators}:quad frac{a}{c}pm frac{b}{c}=frac{apm :b}{c}}
$$
{color{#4257b2}text{c)}}
$$
Solution to this example is given below
$$
begin{align*}
tan left(xright)=0quad mathrm{or}quad sec left(xright)&=0quad mathrm{or}quad sin left(xright)=0&&boxed{text{Solving each part separately}}\
end{align*}
$$
First we solve $tan left(xright)=0$
$$
begin{align*}
x&=0+pi n&&boxed{text{Simplify}}\
x&=color{#c34632}{pi n}&&boxed{text{Simplify}}\
end{align*}
$$
Second we solve $sec left(xright)=0$
$$
begin{align*}
x=operatorname{arcsec} left(0right)+2pi n,:x&=-operatorname{arcsec} left(0right)+2pi n&&boxed{text{General solutions }}\
x&=textbf{ No solution}&&boxed{text{Simplify}}\
end{align*}
$$
Finally we solve $sin left(xright)=0$
$$
begin{align*}
x=0+2pi n,:x&=pi +2pi n&&boxed{text{Simplify}}\
x=2pi n,:x&=pi +2pi n&&boxed{text{Simplify}}\
end{align*}
$$
The equation is $textbf{ Sometimes true}$
$$
begin{align*}
&boxed{{color{#c34632}x=pi n,:x=2pi n,:x=pi +2pi n} }&&boxed{text{Combine all the solutions}}\
end{align*}
$$
{color{#4257b2}text{d)}}
$$
Solution to this example is given below
$$
begin{align*}
sec ^2left(xright)&=0quad mathrm{or}quad cos left(xright)=0&&boxed{text{Solving each part separately}}\
end{align*}
$$
First we solve $sec^2 left(xright)=0$
$$
begin{align*}
sec left(xright)&=0&&boxed{text{Apply rule}}\
x=operatorname{arcsec} left(0right)+2pi n,&:x=-operatorname{arcsec} left(0right)+2pi n&&boxed{text{General solutions}}\
x&=textbf{ No solution}&&boxed{text{Simplify}}\
end{align*}
$$
Second we solve $cos left(xright)=0$
$$
begin{align*}
x&=frac{pi }{2}+2pi n,:x=frac{3pi }{2}+2pi n&&boxed{text{General solutions }}\
end{align*}
$$
The equation is $textbf{ Never true}$
$$
begin{align*}
&boxed{{color{#c34632}text{ No solution for }x inmathbb{R} } }&&boxed{text{Final solution}}\
end{align*}
$$
{color{#4257b2}text{e)}}
$$
Solution to this example is given below
Manipulating left side
$$
begin{align*}
&cos left(xright)sec left(xright)&&boxed{text{Simplify}}\
&cos left(xright)frac{1}{cos left(xright)}&&boxed{text{Using the basic trigonometric identity}}\
&frac{cos left(xright)}{cos left(xright)}&&boxed{text{Multiply fractions}}\
&1&&boxed{text{Cancel the common factor }cos(x)}\
&text{ True}&&boxed{text{Simplify}}\
end{align*}
$$
The equation is $textbf{ Sometimes true}$
$$
begin{align*}
&boxed{{color{#c34632}text{True}} }&&boxed{text{Final solution}}\
end{align*}
$$
color{#4257b2} text{ a) }x=frac{pi }{3}+2pi n,:x=frac{2pi }{3}+2pi n
$$
$$
color{#4257b2} text{ b) } x=frac{pi }{3}+2pi n,:x=frac{5pi }{3}+2pi n
$$
$$
color{#4257b2} text{ c) }x=pi n,:x=2pi n,:x=pi +2pi n
$$
$$
color{#4257b2} text{ d) No solution for }x in mathbb{R}
$$
$$
color{#4257b2} text{ e) True }
$$
Reciprocal of a trigonometric function is meant by dividing the trigonometric function by one. After that, we will sketch the graph of the root function and reciprocal function. The comparison between the graph based on their values at some defined radians will explain what is meant by the term “Reciprocal Trigonometric function”.
$$
begin{align*}
y &= sin(x)tag{1}\
y & = cos(x)tag{2}\
end{align*}
$$
Table for $y = sin(x)$.
begin{center}
renewcommand{arraystretch}{2}
begin{tabular}{ c | c | c | c | c | c | c }
$$ & 0 & $dfrac{pi}{6}$ & $dfrac{pi}{4}$ & $dfrac{pi}{3}$ & $dfrac{pi}{2}$ &$pi$ \
hline
$sin(x)$ & 0 & $dfrac{1}{2}$ & $dfrac{1}{sqrt2}$ & $dfrac{sqrt3}{2}$& $1$ & $0$ \
end{tabular}
end{center}
Table for $y = cos(x)$.\
begin{center}
renewcommand{arraystretch}{2}
begin{tabular}{ c | c | c | c | c | c | c }
$$ & 0 & $dfrac{pi}{6}$ & $dfrac{pi}{4}$ & $dfrac{pi}{3}$ & $dfrac{pi}{2}$ &$pi$ \
hline
$cos(x)$ & $1$ & $dfrac{sqrt3}{2}$ & $dfrac{1}{sqrt2}$ & $dfrac{1}{2}$& $0$ & $-1$\
end{tabular}
end{center}
$$
begin{align*}
y &= dfrac{1}{sin(x)}tag{3} \\
y & = dfrac{1}{cos(x)}tag{4} \\
end{align*}
$$
Table for $y = dfrac{1}{sin(x)}$.
begin{center}
renewcommand{arraystretch}{2}
begin{tabular}{ c | c | c | c | c | c | c }
$$ & 0 & $dfrac{pi}{6}$ & $dfrac{pi}{4}$ & $dfrac{pi}{3}$ & $dfrac{pi}{2}$ &$pi$\
hline
$dfrac{1}{sin(x)}$ & 0 & $2$ & $sqrt2$ & $dfrac{2}{sqrt3}$ & $1$ & $0$\
end{tabular}
end{center}
Table for $y = cos(x)$.\
begin{center}
renewcommand{arraystretch}{2}
begin{tabular}{ c | c | c | c | c | c | c }
$$ & 0 & $dfrac{pi}{6}$ & $dfrac{pi}{4}$ & $dfrac{pi}{3}$ & $dfrac{pi}{2}$ &$pi$\
hline
$dfrac{1}{cos(x)}$ & $1$ & $dfrac{2}{sqrt3}$ & $sqrt2$ & $2$& $0$ & $-1$\
end{tabular}
end{center}
a)
$$
sin (A) = frac {text{side opposite}}{text{hypotenuse}} = frac ac
$$
$$
color{#c34632}{csc (theta) = frac {1}{sin theta}}
$$
Then
$$
csc (A) = frac {1}{sin (A)} = frac {1}{frac ac} = frac ca
$$
$$
cos (B) = frac {text{side adjacent}}{text{hypotenuse}} = frac ac
$$
$$
color{#c34632}{ sec theta = frac {1}{ cos theta}}
$$
Then
$$
sec (B) = frac {1}{cos (B)} = frac {1}{frac ac} = frac ca
$$
b) $csc (A) = frac ca$
c) $cos (B) = frac ac$
d) $sec (B) = frac ca$
$$
4 sin^2 (x) = 1
$$
We can begin by first simplifying the equation,
$$
begin{align*}
4 sin^2 (x) &= 1 \
sin^2 (x) &= dfrac{1}{4} tag{Dividing both sides by 4} \\
sqrt{sin^2 (x)} &= sqrt{dfrac{1}{4}} tag{Square rooting both sides}\\
sin (x) &= pm dfrac{1}{2} \
end{align*}
$$
$sin x = dfrac{1}{2}$, and $sin x = -dfrac{1}{2}$
First we will solve the first solution,
$$
begin{align*}
sin x &= dfrac{1}{2} \\
sin ^{-1} (sin x) &= sin ^{-1} dfrac{1}{2} tag {Taking $sin ^{-1}$ on both sides} \\
x&= sin ^{-1} dfrac{1}{2} \\
x&= dfrac{pi}{6}
end{align*}
$$
We now have x = $dfrac{pi}{6}$.
We also know that $sin$ is positive in second quadrant as well.
This implies $sin theta = sin (pi – theta)$
Therefore,
$sin dfrac{pi}{6} = sin (pi – dfrac{pi}{6}) = sin dfrac{5pi}{6}$
Since we are given 0 $leq$ x $<$ $2 pi$ we have x = $dfrac{pi}{6}$ and x = $dfrac{5pi}{6}$ for the first solution i.e. when $sin x = dfrac{1}{2}$
$$
begin{align*}
sin x &= -dfrac{1}{2} \\
sin ^{-1} (sin x) &= sin ^{-1} left(-dfrac{1}{2}right) tag {Taking $sin ^{-1}$ on both sides} \\
x&= sin ^{-1} left(-dfrac{1}{2}right) \\
x&= -dfrac{pi}{6}
end{align*}
$$
We now have x = $-dfrac{pi}{6}$ but we are given we are given 0 $leq$ x $<$ $2 pi$.
We also know that $sin$ is negative in third and fourth quadrant.
This implies $sin (pi + theta) = -sin theta$
$$
begin{align*}
sin (pi + dfrac{pi}{6}) &= – sin dfrac{pi}{6} \\
sin ( dfrac{7pi}{6}) & = – dfrac{1}{2} tag{as $sin dfrac{pi}{6} = dfrac{1}{2}$}
end{align*}
$$
Also, $sin (2pi – theta) = -sin theta$
$$
begin{align*}
sin (2pi – dfrac{pi}{6}) &= – sin dfrac{pi}{6} \\
sin ( dfrac{11pi}{6}) & = – dfrac{1}{2} tag{as $sin dfrac{pi}{6} = dfrac{1}{2}$} \
end{align*}
$$
Thus we have x = $dfrac{7pi}{6}$, and x = $dfrac{11pi}{6}$ for the second solution i.e. $sin x = -dfrac{1}{2}$
Therefore, for $4 sin^2 (x) = 1$ we have x=$dfrac{pi}{6}$, x=$dfrac{5pi}{6}$, x=$dfrac{7pi}{6}$, x=$dfrac{11pi}{6}$ for
0 $leq$ x $<$ $2 pi$.
$$
3 tan^2 (x) = 1
$$
We can begin by first simplifying the equation,
$$
begin{align*}
3 tan^2 (x) &= 1 \
tan^2 (x) &= dfrac{1}{3} tag{Dividing both sides by 3} \\
sqrt{tan^2 (x)} &= sqrt{dfrac{1}{3}} tag{Square rooting both sides}\\
tan (x) &= pm dfrac{1}{sqrt3} \
end{align*}
$$
$tan x = dfrac{1}{sqrt3}$, and $tan x = -dfrac{1}{sqrt3}$
First we will solve the first solution,
$$
begin{align*}
tan x &= dfrac{1}{sqrt3} \\
tan ^{-1} (tan x) &= tan ^{-1} dfrac{1}{sqrt3} tag {Taking $tan ^{-1}$ on both sides} \\
x&= tan ^{-1} dfrac{1}{sqrt3} \\
x&= dfrac{pi}{6}
end{align*}
$$
We now have x = $dfrac{pi}{6}$.
We also know that $tan$ is positive in the third quadrant as well.
This implies $tan theta = tan (pi + theta)$
Therefore,
$tan dfrac{pi}{6} = tan (pi + dfrac{pi}{6}) = tan dfrac{7pi}{6}$
Since we are given 0 $leq$ x $<$ $2 pi$ we have x = $dfrac{pi}{6}$ and x = $dfrac{7pi}{6}$ for the first solution i.e. when $tan x = dfrac{1}{sqrt3}$
$$
begin{align*}
tan x &= -dfrac{1}{sqrt3} \\
tan ^{-1} (tan x) &= tan ^{-1} left(-dfrac{1}{sqrt3}right) tag {Taking $tan ^{-1}$ on both sides} \\
x&= tan ^{-1} left(-dfrac{1}{sqrt3}right) \\
x&= -dfrac{pi}{6}
end{align*}
$$
We now have x = $-dfrac{pi}{6}$ but we are given we are given 0 $leq$ x $<$ $2 pi$.
We also know that $tan$ is negative in second and fourth quadrant.
This implies $tan (pi – theta) = -tan theta$
$$
begin{align*}
tan (pi – dfrac{pi}{6}) &= – tan dfrac{pi}{6} \\
tan ( dfrac{5pi}{6}) & = – dfrac{1}{sqrt3} tag{as $tan dfrac{pi}{6} = dfrac{1}{sqrt3}$}
end{align*}
$$
Also, $tan (2pi – theta) = -tan theta$
$$
begin{align*}
tan (2pi – dfrac{pi}{6}) &= – tan dfrac{pi}{6} \\
tan ( dfrac{11pi}{6}) & = – dfrac{1}{sqrt3} tag{as $tan dfrac{pi}{6} = dfrac{1}{sqrt3}$} \
end{align*}
$$
Thus we have x = $dfrac{5pi}{6}$, and x = $dfrac{11pi}{6}$ for the second solution i.e. $tan x = -dfrac{1}{sqrt3}$
Therefore, for $3 tan^2 (x) = 1$ we have x=$dfrac{pi}{6}$, x=$dfrac{7pi}{6}$, x=$dfrac{5pi}{6}$, x=$dfrac{11pi}{6}$ for
0 $leq$ x $<$ $2 pi$.
$$
begin{align*}
2sin x cos x+ cos x&=0\
cos x( 2sin x + 1) &= 0\
cos x=0 qquad 2sin x+1&=0\
2sin x&=-1\
sin x &=- frac 12
end{align*}
$$
By applying the inverse function we get
$$
begin{align*}
x&= arccos 0\
x&= frac{pi}2 qquad x=frac{3pi}2\
x&=90^{text{textdegree}} qquad x=270^{text{textdegree}}
end{align*}
$$
and
$$
begin{align*}
x&= arcsin left(-frac 12 right)\
x&= frac{7pi}6 qquad x=frac{11pi}6\
x&=210^{text{textdegree}} qquad x=330^{text{textdegree}}
end{align*}
$$
$$
begin{align*}
x &= frac {pi}{2} + 2k pi qquad x= frac {3 pi}{2} + 2k pi\
x &= 90^{text{textdegree}} + 360^{text{textdegree}} k qquad x= 270^{text{textdegree}} + 360^{text{textdegree}} k
end{align*}
$$
and
$$
begin{align*}
x &= frac {7 pi}{6} + 2k pi qquad x = frac {11 pi}{6} + 2k pi\
x &= 210^{text{textdegree}} + 360^{text{textdegree}} k qquad x = 330^{text{textdegree}} + 360^{text{textdegree}} k
end{align*}
$$
begin{align*}
x &= frac {pi}{2} + 2k pi qquad x= frac {3 pi}{2} + 2k pi\
x &= 90^{text{textdegree}} + 360^{text{textdegree}} k qquad x= 270^{text{textdegree}} + 360^{text{textdegree}} k
end{align*}
$$
and
$$
begin{align*}
x &= frac {7 pi}{6} + 2k pi qquad x = frac {11 pi}{6} + 2k pi\
x &= 210^{text{textdegree}} + 360^{text{textdegree}} k qquad x = 330^{text{textdegree}} + 360^{text{textdegree}} k
end{align*}
$$
2sin xcos x+cos x=0
$$
cos x(2sin x+1)=0
$$
or
$2sin x+1=0$
$2sin x=-1$
$sin x=-dfrac{1}{2}$
$x_2=dfrac{7pi}{6}+2kpi=210text{textdegree}+360text{textdegree} k$
$x_3=dfrac{11pi}{6}+2kpi=330text{textdegree}+360text{textdegree} k$
$x_2=dfrac{7pi}{6}+2kpi=210text{textdegree}+360text{textdegree} k$
$x_3=dfrac{11pi}{6}+2kpi=330text{textdegree}+360text{textdegree} k,k$ integer
a)
$$
begin{align*}
frac {3}{(x-4)(x+1)} + frac {6}{x+1} &= frac {3+6(x-4)}{(x-4)(x+1)}\
&= frac {3+6x-24}{x^2 – 4x +x -4}\
&= frac {6x-21}{x^2 – 3x -4}
end{align*}
$$
Therefore
$$
frac {3}{(x-4)(x+1)} + frac {6}{x+1} = frac {6x-21}{x^2 – 3x -4}
$$
$$
begin{align*}
frac {x+2}{x^2-9} – frac {1}{x+3} &= frac {x+2}{(x-3)(x+3)} – frac {1}{x+3}\
&= frac {x+2-1 cdot (x-3)}{(x-3)(x+3)}\
&= frac {x+2-x+3}{x^2 – 9}\
&= frac {5}{x^2 – 9}
end{align*}
$$
Using the formula
$$
color{#c34632}{a^2+b^2=(a-b)(a+b)}
$$
Therefore
$$
frac {x+2}{x^2-9} – frac {1}{x+3} = frac {5}{x^2 – 9}
$$
b) $frac {x+2}{x^2-9} – frac {1}{x+3} = frac {5}{x^2 – 9}$
$y=x-5 qquad rightarrow qquad x-y=5$ (2)
$x=2z qquad rightarrow qquad x-2z=0$ (3)
Adding (1) and (2) to eliminate $y$
$x+y+z=40$ (1)
+
$x-y=5$ (2)
————————
$2x+z=45$
$4x+2z=90$ (Multiply (4) by 2.
Adding (3) and (4)
$x-2z=0$ (3)
+
$4x+2z=90$ (4)
—————————
$5x=90$
$$
x=18
$$
Substituting 18 for $x$ in equations (2) and (3)
$y=18-5$
$$
y=13
$$
$18=2z$
$$
z=9
$$
The solution is: $(18, 13, 9)$
$$
f(x)= frac 2{x-1}+3
$$
$$
g(x)= -0.5(x+2)^3-3
$$
$$
h(x)=left|x+5 right|-4
$$
$$
k(x)= 2sqrt{x+1}+3
$$
f(x)=dfrac{2}{x-1}+3
$$
g(x)=-0.5(x+2)^3-3
$$
h(x)=|x+5|-4
$$
k(x)=2sqrt{x+1}+3
$$
(a-b)^8
$$
T_{k+1}=_nC_k a^{n-k}b^k
$$
$n-k=5$
$$
Rightarrow k=3
$$
$=-dfrac{cancel{5!}cdot 6cdot 7cdot 8}{cancel{5!}cdot 1cdot 2cdot 3}cdot a^5b^3$
$$
=-56a^5b^3
$$
T_4=-56a^5b^3
$$
a) Since
$$
begin{align*}
sin (A) &= frac ac\
cos (A) &= frac bc
end{align*}
$$
according to the formula
$$
color{#c34632}{tan (theta) = frac {sin (theta)}{cos (theta)}}
$$
we obtain
$$
tan (A) = frac {sin (A)}{cos (A)} = frac {frac ac}{frac bc} = frac ab
$$
$$
begin{align*}
sin (A) &= frac ac\
cos (A) &= frac bc
end{align*}
$$
according to the formula
$$
color{#c34632}{cot (theta) = frac {cos (theta)}{sin (theta)}}
$$
we obtain
$$
cot (A) = frac {cos (A)}{sin (A)} = frac {frac bc}{frac ac} = frac ba
$$
$$
begin{align*}
sin (B) &= frac bc\
cos (B) &= frac ac
end{align*}
$$
according to the formula
$$
color{#c34632}{tan (theta) = frac {sin (theta)}{cos (theta)}}
$$
we obtain
$$
tan (B) = frac {sin (B)}{cos (B)} = frac {frac bc}{frac ac} = frac ba
$$
b) $cot (A) = frac ba$
c) $tan (B) = frac ba$
y=2sinleft(dfrac{x}{2}+dfrac{3pi}{4}right)+1
$$
$$
a=2
$$
$$
T=4pi
$$
y=5cosleft(x-dfrac{pi}{2}right)-2
$$
$$
a=5
$$
$$
T=2pi
$$
b) $y=5cosleft(x-dfrac{pi}{2}right)-2$
$r+g+b=40$ (The sum of marbles are 40 mrbles) (1)
$r=b+5$ (5 more marbles than blue) (2)
$r=2g$ (Red marbles are twice the green marbles) (3)
Substituting for $r$ from equation (3) in equations (1) and (2)
$2g+g+b=40$
$3g+b=40$ (1′)
$2g=b+5$
$2g-b=5$ (2′)
Adding (1′) and (2′)
$3g+b=40$ (1′)
+
$2g-b=5$ (2′)
———————–
$5g qquad =45$
$$
g=9
$$
Substituting in equation (3)
$r=2(9)$ (3)
$$
r=18
$$
Substituting in equation (1)
$18+9+b=40$
$b=40-27$
$$
b=13
$$
The number of each color of marbles is:
18 red marbles.
9 green marbles.
13 blue marbles.
180text{textdegree}-6text{textdegree}-63text{textdegree}=111text{textdegree}
$$
$xsin 111text{textdegree}=10sin 6text{textdegree}$
$$
x=dfrac{10cdot 0.10452846}{0.93358043}approx 1.12
$$
$=100+4900-1400cdot0.64278761$
$=4100.0973$
$x=sqrt{4100.0973}$
$$
xapprox 64.03
$$
b) 64.03
$approx 0.081$ peaches/cubic inch
$0.081peaches/inch^3=dfrac{0.081 peaches}{left(dfrac{1}{12}right)^3foot^3}$
$$
approx 140peaches/foot^3
$$
$$
begin{align*}
color{#c34632}{(a+b)^2 = a^2 + 2ab + b^2}\
color{#c34632}{(a-b)^2 = a^2 – 2ab + b^2}
end{align*}
$$
Therefore
$$
begin{align*}
(x+2)^2 – (x-2)^2 &= 8\
(x^2 + 4x +4) – ( x^2 – 4x + 4) &= 8\
x^2 + 4x + 4 – x^2 +4x -4 &= 8\
8x &= 8\
x &=1
end{align*}
$$
The solution of equation is
$$
x=1
$$
$$
begin{align*}
frac 1x + frac {1}{x+2} &= 3\
frac {x+2+x}{x(x+2)} &= 3\
frac {2x+2}{x(x+2)} &= 3\
2x+2 &= 3x(x+2)\
2x+2 &= 3x^2 + 6x\
3x^2 + 6x – 2x-2 &= 0\
3x^2 + 4x – 2 &= 0
end{align*}
$$
The solution of a square equation $ax^2 + bx+c=0$ is shape
$$
color{#c34632}{x= frac {-b pm sqrt {b^2 – 4ac}}{2a}}
$$
Now, we have
$$
begin{align*}
x &= frac {-4 pm sqrt {16 + 24}}{2 cdot 3}\
x &= frac {-4 pm sqrt {40}}{6}\
x &= frac {-4 pm 2 sqrt {10}}{6}\
x &= frac {-2 pm sqrt {10}}{3}
end{align*}
$$
The solution of equation is
$$
x = frac {-2 + sqrt {10}}{3} qquad text{and} qquad x = frac {-2 – sqrt {10}}{3}
$$
$$
begin{align*}
11(4^x) -15 &= -14\
11(4^x) &= 1\
4^x &= frac {1}{11}
end{align*}
$$
We use the $ln$ function on both sides of the equation
$$
ln 4^x = ln frac {1}{11}
$$
According to the formula
$$
color{#c34632}{ln a^b = b ln a}
$$
we have
$$
begin{align*}
x ln 4 &= ln frac {1}{11}\
x &= frac {ln frac {1}{11}}{ln 4}
end{align*}
$$
The solution of equation is
$$
x = frac {ln frac {1}{11}}{ln 4}
$$
$$
begin{align*}
4 log_{13}{x} – 8 &= 16\
4 log_{13}{x} &= 24\
log_{13}{x} &= 6
end{align*}
$$
According to the formula
$$
begin{align*}
color{#c34632}{log_ba = c}\
color{#c34632}{a = b^c}
end{align*}
$$
we have
$$
x= 13^6
$$
The solution of equation is
$$
x= 13^6
$$
b) $x = frac {-2 + sqrt {10}}{3} qquad text{and} qquad x = frac {-2 – sqrt {10}}{3}$
c) $x = frac {ln frac {1}{11}}{ln 4}$
d) $x=13^6$
(x+2)^2-(x-2)^2=8
$$
$8x=8$
$$
x=1
$$
dfrac{1}{x}+dfrac{1}{x+2}=3
$$
(-infty,-2)cup(-2,0)cup(0,infty)
$$
$x+2+x=3x(x+2)$
$$
2x+2=3x^2+6x
$$
$$
3x^2+4x-2=0
$$
$=dfrac{-4pm 2sqrt{10}}{6}$
$=dfrac{-2pmsqrt{10}}{3}$
$x_1=dfrac{-2-sqrt{10}}{3}$
$x_2=dfrac{-2+sqrt{10}}{3}$
11(4^x)-15=-14
$$
$$
11(4^x)=1
$$
4^x=dfrac{1}{11}
$$
$xln 4=lndfrac{1}{11}$
$x=dfrac{lndfrac{1}{11}}{ln 4}$
$$
xapprox -1.73
$$
4log_{13} x-8=16
$$
$$
4log_{13} x=24
$$
log_{13} x=6
$$
x=13^6=4,826,809
$$
$$2 sin (x) cos (x) = dfrac{1}{2}$$ \
a. We have to find if we can solve for x algebraically. We can do this by simplifying the equation. \
begin{align*}
2 sin (x) cos (x) &= dfrac{1}{2} \\
sin (x) cos (x) + sin (x) cos (x) &= dfrac{1}{2} \\
intertext{Using the identity, $sin A cos B + sin B cos A = sin (A+B)$, we get} \
sin (x + x) &= dfrac{1}{2} \\
sin (2x) & = dfrac{1}{2} \\
end{align*}
Now we have $sin (2x) = dfrac{1}{2} $.
begin{align*}
sin (2x) & = dfrac{1}{2} \\
intertext{Taking $sin ^{-1}$ on both sides} \
sin ^{-1} [sin (2x)] & = sin ^{-1} left(dfrac{1}{2}right) \\
2x &= dfrac{pi}{6} tag{$sin ^{-1} left(dfrac{1}{2}right)= dfrac{pi}{6}$}\\
x&= dfrac{pi}{12} \\
end{align*}
We also know $sin theta$ is positive in second quadrant as well. \
Therefore, we can say $sin theta = sin (pi – theta)$ \
Accordingly $sin 2x = sin (pi – 2x)$ \
Also, $2x = dfrac{pi}{6}$ \
Thus, $sin 2x = sin (pi – dfrac{pi}{6})$ or $sin 2x = sin dfrac{5pi}{6}$ \
Therefore, we have $2x=dfrac{5pi}{6}$. \\
Thus, $x=dfrac{5pi}{12}$.\
We are given $0leq x < 2 pi$ \
And in this interval we have $x=dfrac{pi}{12}$ and $x=dfrac{5pi}{12}$ for $sin (2x) = dfrac{1}{2}$ \
Therefore, for $sin (2x) = dfrac{1}{2}$ we had found $x=dfrac{pi}{12}$ and $x=dfrac{5pi}{12}$
Additionally, $x=dfrac{pi}{12} + pi$ or $x=dfrac{13pi}{12}$
Similarly, $x=dfrac{5pi}{12} + pi$ and $x=dfrac{17pi}{12}$
Therefore, we have 4 solutions in our interval $0leq x < 2 pi$
$x=dfrac{pi}{12}$
$x=dfrac{5pi}{12}$
$x=dfrac{13pi}{12}$
$$
x=dfrac{17pi}{12}
$$
$$
x= dfrac{pi}{12}, dfrac{5pi}{12}, dfrac{13pi}{12}, dfrac{17pi}{12}
$$
begin{align*}
2 sin (x) cos (x) &= sin (x) cos (x) + sin (x) cos (x) \\
intertext{Using the identity, $sin A cos B + sin B cos A = sin (A+B)$, we get} \
2 sin (x) cos (x) &= sin (x + x) \\
&=sin (2x) \\
end{align*}
Thus, $2 sin (x) cos (x) = sin 2x $
As we can observe the graph is exactly the same of both the equations. Thus there is more than 1 function. The function $sin 2x = dfrac{1}{2}$ is the simplest.
begin{align*}
sin (2x) & = dfrac{1}{2} \\
intertext{Taking $sin ^{-1}$ on both sides} \
sin ^{-1} [sin (2x)] & = sin ^{-1} left(dfrac{1}{2}right) \\
2x &= dfrac{pi}{6} tag{$sin ^{-1} left(dfrac{1}{2}right)= dfrac{pi}{6}$}\\
x&= dfrac{pi}{12} \\
end{align*}
We also know $sin theta$ is positive in second quadrant as well. \
Therefore, we can say $sin theta = sin (pi – theta)$ \
Accordingly $sin 2x = sin (pi – 2x)$ \
Also, $2x = dfrac{pi}{6}$ \
Thus, $sin 2x = sin (pi – dfrac{pi}{6})$ or $sin 2x = sin dfrac{5pi}{6}$ \
Therefore, we have $2x=dfrac{5pi}{6}$. \\
Thus, $x=dfrac{5pi}{12}$.\
We are given $0leq x < 2 pi$ \
And in this interval we have $x=dfrac{pi}{12}$ and $x=dfrac{5pi}{12}$ for $sin (2x) = dfrac{1}{2}$ \
Therefore, for $sin (2x) = dfrac{1}{2}$ we had found $x=dfrac{pi}{12}$ and $x=dfrac{5pi}{12}$
Additionally, $x=dfrac{pi}{12} + pi$ or $x=dfrac{13pi}{12}$
Similarly, $x=dfrac{5pi}{12} + pi$ and $x=dfrac{17pi}{12}$
Therefore, we have 4 solutions in our interval $0leq x < 2 pi$
$x=dfrac{pi}{12}$
$x=dfrac{5pi}{12}$
$x=dfrac{13pi}{12}$
$x=dfrac{17pi}{12}$
As we can clearly see the values algebraically and from the graph are the same.
$$
2 sin (x) cos (x) = sin 2x
$$
And this is the Double Angle Identity for sine.
a. $y=cos (-x)$
We will first graph this function.
So we can write $y=cos (-x)$ as $y=sin [90text{textdegree}-(-x)]$ or $y=sin (90text{textdegree}+x)$
We also know $sin (A+B) = sin A cos B + sin B cos A$
We will now simplify the equation,
$$
begin{align*}
sin (90text{textdegree}+x) &= sin 90text{textdegree} cos x + cos 90text{textdegree} sin x \
&=(1) times cos x + (0) times sin x tag{$sin 90text{textdegree} = 1$ and $cos 90text{textdegree} = 0$} \
&= cos x \
end{align*}
$$
Therefore we get $sin (90text{textdegree}+x) = cos x$
This implies, $cos(-x)=cos(x)$.
We can also check by drawing the graph of $y=cos x$ and comparing it with the graph of $y = cos (-x)$ drawn above.
We can clearly see both the graphs are exactly the same.
Therefore, we have a new identity
$$
boxed {cos(-x)=cos(x)}
$$
We will first graph this function.
So we can write $y=sin (-x)$ as $y=cos [90text{textdegree}-(-x)]$ or $y=cos (90text{textdegree}+x)$
We also know $cos (A+B) = cos A cos B – cos B cos A$
We will now simplify the equation,
$$
begin{align*}
cos (90text{textdegree}+x) &= cos 90text{textdegree} cos x – sin 90text{textdegree} sin x \
&=(0) times cos x – (1) times sin x tag{$sin 90text{textdegree} = 1$ and $cos 90text{textdegree} = 0$} \
&= – sin x \
end{align*}
$$
Therefore we get $cos (90text{textdegree}+x) = -sin x$
This implies, $sin (-x)=-sin(x)$.
We can also check by drawing the graph of $y=-sin x$ and comparing it with the graph of $y = sin (-x)$ drawn above.
We can clearly see both the graphs are exactly the same.
Therefore, we have a new identity
$$
boxed {sin(-x)=-sin(x)}
$$
In terms of degree, this can be written as $y = cos (90- x)$
We will first draw the graph of this function.
We will now simplify the equation,
$$
begin{align*}
cos (90text{textdegree}-x) &= cos 90text{textdegree} cos x + sin 90text{textdegree} sin x \
&=(0) times cos x+ (1) times sin x tag{$sin 90text{textdegree} = 1$ and $cos 90text{textdegree} = 0$} \
&= sin x \
end{align*}
$$
Therefore we get $cos (90text{textdegree}-x) = sin x$
We can also check by drawing the graph of $y=sin x$ and comparing it with the graph of $y =cos (90text{textdegree}-x)$ drawn above.
We can clearly see both the graphs are very similar.
Therefore, we have the identity
$$
boxed {cos (dfrac{pi}{2} – x)=sin(x)}
$$
In terms of degree, this can be written as $y = sin (90- x)$
We will first draw the graph of this function.
We will now simplify the equation,
$$
begin{align*}
sin (90text{textdegree}-x) &= sin 90text{textdegree} cos x – sin x cos 90text{textdegree} \
&=(1) times cos x+ sin x times (0) tag{$sin 90text{textdegree} = 1$ and $cos 90text{textdegree} = 0$} \
&= cos x \
end{align*}
$$
Therefore we get $sin (90text{textdegree}-x) = cos x$
We can also check by drawing the graph of $y=cos x$ and comparing it with the graph of $y =sin (90text{textdegree}-x)$ drawn above.
We can clearly see both the graphs are very similar.
Therefore, we have the identity
$$
boxed {sin (dfrac{pi}{2} – x)=cos(x)}
$$
We will first graph this function.
Also, $cos (A+B) = cos A cos B – sin A sin B$ \
By using this identity we can find the value of $cos(2x)$ \
begin{align*}
cos (2x) &= cos(x + x) \
&= cos x cos x – sin x sin x \
&= cos^2 x – sin^2 x \
intertext{We know $sin^2 x + cos^2 x = 1$ or $sin^2 x =1 – cos^2 x $}
cos (2x) &= cos^ 2x – (1 – cos^2 x) \
&= cos^ 2x – 1 + cos^2 x \
&= 2 cos^ 2x – 1 \
end{align*}
Thus, we get $cos (2x) = 2 cos^ 2x – 1 $ or $cos (2x) + 1 = 2 cos^ 2x $ or \
$dfrac{cos (2x) + 1}{2}= cos^ 2x $
Now we have $cos^ 2x = dfrac{cos (2x) + 1}{2}$ \
We will graph the function $y= dfrac{cos (2x) + 1}{2}$ to check if we have at the right conclusion. \
We can clearly observe that the graph of the function $y=cos^ 2x $ drawn \
above
is identical to graph of the function $y= dfrac{cos (2x) + 1}{2}$ drawn below. \
Therefore, we have the identity $$boxed {cos^ 2x= dfrac{cos (2x) + 1}{2}}$$
begin{align*}
cos (2x) &= cos(x + x) \
intertext{using $cos(A+B) = cos A cos B – sin A sin B$}
&= cos x cos x – sin x sin x \
&= cos^2 x – sin^2 x \
end{align*}
Thus, we get $cos (2x) = cos^2 x – sin^2 x $ or \
$cos (2x) + sin^2 x= cos^ 2x $
Now we have $cos^ 2x = cos (2x) + sin^2 x$ \
We will graph the function $y= cos (2x) + sin^2 x$ to check if we have at the right conclusion. \
We can clearly observe that the graph of the function $y=cos^ 2x $ drawn \
above
is identical to graph of the function $y=cos (2x) + sin^2 x$ drawn below. \
Therefore, we have the identity $$boxed {cos^ 2x=cos (2x) + sin^2 x}$$
We will first graph this function.
Also, $cos (A+B) = cos A cos B – sin A sin B$ \
By using this identity we can find the value of $cos(2x)$ \
begin{align*}
cos (2x) &= cos(x + x) \
&= cos x cos x – sin x sin x \
&= cos^2 x – sin^2 x \
intertext{We know $sin^2 x + cos^2 x = 1$ or $ cos^2x = 1 – sin^2 x$}
cos (2x) &= 1 – sin^2 x – sin^2x \
&= 1 – 2 sin^ 2x \
end{align*}
Thus, we get $cos (2x) = 1 – 2 sin^ 2x $ or $1 – cos (2x) = 2 sin^ 2x $ or \
$dfrac{1 – cos (2x)}{2}= sin^ 2x $
Now we have $cos^ 2x = dfrac{1 – cos (2x)}{2}$ \
We will graph the function $y= dfrac{1 – cos (2x)}{2}$ to check if we have at the right conclusion. \
We can clearly observe that the graph of the function $y=sin^ 2x $ drawn \
above
is identical to graph of the function $y= dfrac{1 – cos (2x)}{2}$ drawn below. \
Therefore, we have the identity $$boxed {sin^ 2x= dfrac{1 – cos (2x)}{2}}$$
begin{align*}
cos (2x) &= cos(x + x) \
intertext{using $cos(A+B) = cos A cos B – sin A sin B$}
&= cos x cos x – sin x sin x \
&= cos^2 x – sin^2 x \
end{align*}
Thus, we get $cos (2x) = cos^2 x – sin^2 x $ or \
$ cos^2 x – cos (2x) = sin^ 2x $
Now we have $sin^ 2x =cos^2 x – cos (2x)$ \
We will graph the function $y= cos^2 x – cos (2x)$ to check if we have at the right conclusion. \
We can clearly observe that the graph of the function $y=sin^ 2x $ drawn \
above
is identical to graph of the function $y=cos^2 x – cos (2x)$ drawn below. \
Therefore, we have the identity $$boxed {sin^ 2x=cos^2 x – cos (2x)}$$
We will first graph this function,
begin{align*}
cos (2x) &= cos(x + x) \
intertext{using $cos(A+B) = cos A cos B – sin A sin B$}
&= cos x cos x – sin x sin x \
&= cos^2 x – sin^2 x \
end{align*}
Thus, we get $ cos^2 x – sin^2 x =cos (2x) $ or \
We will graph the function $y= cos 2x$ to check if we have at the right conclusion. \
We can clearly observe that the graph of the function $y=cos 2x $ drawn \
above
is identical to graph of the function $y= cos^2 x – sin^2 x$ drawn below. \
Therefore, we have the identity $$boxed { cos^2 (x) – sin^2 (x)=cos 2 x }$$
We will first graph this function,
Therefore, we have the identity
$$
boxed { cos^2 (x) + sin^2 (x)=1 }
$$
We can simplify it further, \
begin{align*}
2 cos left(dfrac{pi}{2} – xright) &= 1 \\
cos left(dfrac{pi}{2} – xright) &= dfrac{1}{2} \\
intertext{We have the identity $cos left(dfrac{pi}{2} – x right) = sin x$, this implies }
sin x &= dfrac{1}{2}\\
sin^ {-1} (sin x)&= sin^{-1} left(dfrac{1}{2}right) tag{taking $sin^{-1}$ on both sides} \\
x&= dfrac{pi}{6} tag{$sin^{-1} left(dfrac{1}{2} right) = dfrac{pi}{6}$ }
end{align*}
Therefore, we have $x= dfrac{pi}{6}$ \
Now, we are not given any domain. \
This implies, $sin (pi – theta) = sin theta$
Accordingly,
$$
begin{align*}
sin left(pi – dfrac{pi}{6} right) &= sin dfrac{pi}{6} \\
sin dfrac{5 pi}{6} & = sin dfrac{pi}{6} \\
end{align*}
$$
Therefore we have x = $dfrac{pi}{6}$ and x = $dfrac{5 pi}{6}$
The period of $sin x$ is $2 pi$. Therefore, the value of $sin x = dfrac{1}{2}$ will repeat after adding 2 $pi$ n number of times (where n is a positive integer)
This implies,
x=$dfrac{pi}{6} + 2pi n$ and
x=$dfrac{5pi}{6} + 2pi n$
for $2 cos left(dfrac{pi}{2} – xright) = 1$ (for all x)
$$
cos^2 x – sin^2 x = dfrac{sqrt3}{2}
$$
We have an identity $cos^2 x – sin^2 x = dfrac{sqrt3}{2} = cos (2x)$
Therefore, we have $cos (2x) = dfrac{sqrt3}{2}$
$$
begin{align*}
cos (2x) &= dfrac{sqrt3}{2} \\
cos^{-1} (cos 2x)&= cos^{-1}left( dfrac{sqrt3}{2}right) tag{Taking $cos^{-1} $ on both sides} \\
2x &= dfrac{pi}{6} tag{$cos^{-1}left( dfrac{sqrt3}{2}right) =dfrac{pi}{6} $} \\
x&= dfrac{pi}{12} \\
end{align*}
$$
We have $x= dfrac{pi}{12}$
This implies, $cos (2 pi – theta) = cos theta$
Therefore,
$$
begin{align*}
cos (2 pi – 2x) &= cos 2x \\
cos left(2 pi – dfrac{pi}{6}right)& = cos dfrac{pi}{6} tag{$2x = dfrac{pi}{6}$} \\
cos dfrac{11 pi}{6} &= cos dfrac{pi}{6}
end{align*}
$$
Therefore we have 2x = $dfrac{pi}{6}$ and 2x = $dfrac{11 pi}{6}$
And accordingly, x = $dfrac{pi}{12}$ and x = $dfrac{11 pi}{12}$
This implies,
x=$dfrac{pi}{12} + pi$ or $dfrac{13pi}{12}$ and
x=$dfrac{11pi}{12} + pi$ or $dfrac{23pi}{12}$
Therefore we have
x =$dfrac{pi}{12}$;
x =$dfrac{11pi}{12}$;
x =$dfrac{13pi}{12}$;
x =$dfrac{13pi}{12}$
for $cos^2 x – sin^2 x = dfrac{sqrt3}{2}$ where $0 leq x<2pi$
We will first draw a graph of $x^2 + y^2 = 1$ and a tringle inside it as shown below.
Base = BC = b units
Height = AB = a units
Hypotenuse = AC = c units
By Pythagoras theorem,
$text{(Base)}^2 + text{(Height)}^2 = text{(Hypotenuse)}^2$
Therefore, we have
$$begin{aligned}
text{(BC)}^2 + text{(AB)}^2 &= text{(AC)}^2 \
b^2 + a^2 &= c^2 \
end{aligned}$$
Now we know,
$sin theta = dfrac{text{Height}}{text{Hypotenuse}}$
As per the figure above, we have
$sin x = dfrac{text{AB}}{text{AC}} = dfrac{a}{c}$
Also,
$cos theta = dfrac{text{Base}}{text{Hypotenuse}}$
As per the figure above, we have
$cos x = dfrac{text{BC}}{text{AC}} = dfrac{b}{c}$
and as per the Pythagoras theorem, we have $b^2 + a^2 = c^2$
Now,
$$begin{aligned}
sin^2 x + cos^2 x &= left(dfrac{a}{c}right)^2 + left(dfrac{b}{c}right)^2 \\
&= dfrac{a^2 + b^{2}}{c^{2}} \\
&= dfrac{c^{2}}{c^{2}} \\
&=1 \
end{aligned}$$
As shown above we have derived the identity
$sin^2 x + cos^2 x = 1$ with the help of the Pythagoras theorem. Therefore, this identity is also called the Pythagorean Identity. \
For this, we must keep in mind
$$begin{aligned}
csc theta &= dfrac{1}{sin theta} \\
cot theta &= dfrac{cos theta}{sin theta} \\
tan theta &= dfrac{sin theta}{cos theta} \\
sec theta &= dfrac{1}{cos theta} \\
end{aligned}$$
$$begin{aligned}
sin^2 x + cos^2 x &= 1 \\
dfrac{sin^2 x}{sin^2 x} + dfrac{cos^2 x}{sin^2 x} &= dfrac{1}{sin^2 x} \\
1 + cot^2 x &= csc^2 x
end{aligned}$$
Therefore we have derived a new identity,
$$boxed{1 + cot^2 x = csc^2 x}$$
$$begin{aligned}
sin^2 x + cos^2 x &= 1 \\
dfrac{sin^2 x}{cos^2 x} + dfrac{cos^2 x}{cos^2 x} &= dfrac{1}{cos^2 x} \\
tan^2 x + 1&= sec^2 x
end{aligned}$$
Therefore we have derived a new identity, \\
$$boxed{tan^2 x + 1= sec^2 x}$$
a. The expression is $tan (theta)$
We know, $tan (theta) = dfrac{sin (theta)}{cos (theta)}$
We have the Pythagorean Identity,
$$sin^2 (theta)+cos^2 (theta) =1$$
Thus, we can write$sin (theta)$ in terms of $cos (theta)$ and vice versa.
$$begin{aligned}
sin^2 (theta)+cos^2 (theta) &=1 \
sin^2 (theta)&= 1 – cos^2 (theta) \
sin (theta)&= sqrt{1 – cos^2 (theta)} \
end{aligned}$$
Therefore we can write,
$$boxed{tan (theta)=dfrac{sqrt{1 – cos^2 (theta)}}{cos (theta)}}$$
$$begin{aligned}
sin^2 (theta)+cos^2 (theta) &=1 \
cos^2 (theta)&= 1 – sin^2 (theta) \
cos (theta)&= sqrt{1 – sin^2 (theta)} \
end{aligned}$$
Therefore we can write,
$$boxed{tan (theta)=dfrac{sin (theta)}{sqrt{1 – sin^2 (theta)}}}$$
This can be written as
$$boxed{csc (theta) = dfrac{1}{sin (theta)}}$$
We know, $cot (theta) = dfrac{cos (theta)}{sin (theta)}$
We have the Pythagorean Identity,
$$sin^2 (theta)+cos^2 (theta) =1$$
Thus, we can write$sin (theta)$ in terms of $cos (theta)$ and vice versa.
$$begin{aligned}
sin^2 (theta)+cos^2 (theta) &=1 \
sin^2 (theta)&= 1 – cos^2 (theta) \
sin (theta)&= sqrt{1 – cos^2 (theta)} \
end{aligned}$$
Therefore we can write,
$$boxed{cot (theta)=dfrac{cos (theta)}{sqrt{1 – cos^2 (theta)}}}$$
$$begin{aligned}
sin^2 (theta)+cos^2 (theta) &=1 \
cos^2 (theta)&= 1 – sin^2 (theta) \
cos (theta)&= sqrt{1 – sin^2 (theta)} \
end{aligned}$$
Therefore we can write,
$$boxed{cot (theta)=dfrac{sqrt{1 – sin^2 (theta)}}{sin (theta)}}$$
This can be written as
$$boxed{sec (theta) = dfrac{1}{cos (theta)}}$$
$$
begin{align*}
2 sin (x) &= 1\
sin x &= frac 12
end{align*}
$$
By applying the inverse formula we get
$$
begin{align*}
x &= arcsin frac 12\
x &= frac {pi}{6} qquad x = frac {5 pi}{6}
end{align*}
$$
The solutions of equation are
$$
x = frac {pi}{6} qquad x = frac {5 pi}{6}
$$
$$
cos^2 x + 4 cos x + 4 = 0
$$
Let $cos x=a$ then we have
$$
a^2 + 4a +4=0
$$
The solution of a square equation $ax^2 + bx + c= 0$ is shape
$$
color{#c34632}{x = frac {-b pm sqrt {b^2 – 4ac}}{2a}}
$$
Therefore
$$
begin{align*}
a &= frac {-4 pm sqrt {16 – 16}}{2}\
a &= frac {-4 pm 0}{2}\
a &= -2
end{align*}
$$
Substitute $a= cos x$
$$
cos x = -2
$$
Since
$$
-1 leq cos x leq 1
$$
there is no solution.
b) no solution
2sin x=1
$$
sin x=dfrac{1}{2}
$$
cos^2 x+4cos x+4=0
$$
$$
cos x+2=0
$$
cos x=-2
$$
b) no solution
$$begin{aligned}
text{Period} T&= 81-15- left(dfrac{180}{360+180}right)\\
&= 66- dfrac{180}{504}\\
&= 44\
end{aligned}$$
text{Amplitude} Y&=dfrac{23+|-17|}{2}\\
&=dfrac{40}{2}\\
&=20
end{aligned}$$
$$y = Y cos left(dfrac{2pi x}{T}right)+(23-20)$$
$$y=20 cos left(dfrac{2pi (x-15)}{T}right)+3 $$
$$y(81)= 20 cos(3 pi) +3= -17$$
$$
frac{x^2}{x-5}-frac{25}{x-5}=frac{x^2-25}{x-5}
$$
Now we use the following rule of the difference of squares:
$$
x^2-y^2 = (x-y)cdot (x+y)
$$
and apply it to our problem.
$$
frac{x^2-25}{x-5} = frac{(x-5)cdot(x+5)}{x-5}=x+5
$$
$$
frac{a^2}{a+5}+frac{10a + 25}{a+5} = frac{a^2 + 10a + 25}{a+5}
$$
We can recognize the square of the sum:
$$
(a+b)^2= a^2+2ab+b^2
$$
and apply it to our problem.
$$
frac{a^2 + 10a + 25}{a+5} = frac{(a+5)^2}{a+5} = a+5
$$
$$
frac{x^2}{x-y}-frac{2xy-y^2}{x-y}=frac{x^2-(2xy-y^2)}{x-y}=frac{x^2-2xy+y^2}{x-y}
$$
Just like in the task before we use the square rule only this time it’s the square of the difference:
$$
(a-b)^2= a^2-2ab+b^2
$$
in our problem now it looks like:
$$
frac{x^2-2xy+y^2}{x-y}=frac{(x-y)^2}{x-y} = x-y
$$
$$
frac{x}{x+1}+frac{1}{x-1} = frac{x(x-1)+1(x+1)}{(x-1)(x+1)}=
$$
$$
frac{x^2+x-x+1}{(x-1)(x+1)} = frac{x^2+1}{x^2-1}
$$
$w=c-5$ (1)
$c=2p$ (2)
$w+c+p=40$ (3)
Substituting for $c$ from equation (2) in equations (1) and (3)
$w=2p-5$
$w-2p=-5$ (1′)
$w+3p=40$ (2′)
Subtracting (1′) from (2′)
$w+3p=40$ (2′)
–
$w-2p=-5$ (1′)
————————-
$0+5p=45$
$$
p=9
$$
Substituting in equation (2)
$c=2(9)$
$$
c=18
$$
Substituting in equation (1)
$w=2(9)-5=18-5$
$$
w=13
$$
Pennant from Washington campus = 13
Pennant from California campus = 18
Pennant from Pennsylvania campus = 9
Pennant from California campus = 18
Pennant from Pennsylvania campus = 9
begin{cases}
x+y+z=40\
y=x-5\
x=2z
end{cases}
$$
$$
a+ar+ar^2 +cdot cdot cdot
$$
Therefore
$$
8-6+ frac 92 – frac {27}{8} + cdot cdot cdot
$$
where
$$
a=8 qquad text{and} qquad r = – frac 34
$$
According to the formula
$$
color{#c34632}{S = frac {a}{1-r} }
$$
we obtain
$$
begin{align*}
S &= frac {8}{1- left( – frac 34 right)}\
S &= frac {8}{1+ frac 34}\
S &= frac {8}{frac 74}\
S &= frac {32}{7}
end{align*}
$$
The sum is
$$
S = frac {32}{7}
$$
$$
sum_{k=0}^{infty} frac 18 2^k = frac 18 + frac 14 + frac 12 +cdot cdot cdot
$$
where
$$
a= frac 18 qquad text{and} qquad r=2
$$
According to the formula
$$
color{#c34632}{S = frac {a}{1-2}}
$$
we obtain
$$
begin{align*}
S &= frac {frac 18}{1-2}\
S &= frac {frac 18}{-1} = – frac 18
end{align*}
$$
The sum is
$$
S = – frac 18
$$
b) $S = – frac 18$
8-6+dfrac{9}{2}-dfrac{27}{8}+…
$$
$a_{n}=-dfrac{6}{8}a_{n-1}=-dfrac{3}{4}a_{n-1}$
$$
q=-dfrac{3}{4}
$$
$=dfrac{8}{dfrac{7}{4}}$
$=8cdotdfrac{4}{7}$
$$
=dfrac{32}{7}
$$
sum_{k=0}^{infty}dfrac{1}{8}(2)^k
$$
$a_n=2a_{n-1}$
$$
q=2
$$
b) divergent series
$$cos^2 (x) + sin^2 (x) = 1$$
We can rewrite it in various ways.
For this, we must keep in mind \
$$begin{aligned}
csc theta &= dfrac{1}{sin theta} \\
cot theta &= dfrac{cos theta}{sin theta} \\
tan theta &= dfrac{sin theta}{cos theta} \\
sec theta &= dfrac{1}{cos theta} \\
end{aligned}$$
$$begin{aligned}
sin^2 x + cos^2 x &= 1 \\
dfrac{sin^2 x}{sin^2 x} + dfrac{cos^2 x}{sin^2 x} &= dfrac{1}{sin^2 x} \\
1 + cot^2 x &= csc^2 x
end{aligned}$$
Therefore we have derived a new identity, \\
$$boxed{1 + cot^2 x = csc^2 x}$$
$$begin{aligned}
sin^2 x + cos^2 x &= 1 \\
dfrac{sin^2 x}{cos^2 x} + dfrac{cos^2 x}{cos^2 x} &= dfrac{1}{cos^2 x} \\
tan^2 x + 1&= sec^2 x \
end{aligned}$$
Therefore we have derived a new representation, \\
$$boxed{tan^2 x + 1= sec^2 x} $$
This can also be written as $2 cos^2 (x) = cos (2x)+1$ or
$cos^2 (x) = dfrac{cos (2x)+1}{2}$
Substituting the value of $cos^2 (x)$ in the Pythagorean identity, we get
$$begin{aligned}
sin^2 (x) + dfrac{cos (2x)+1}{2} &= 1 \\
dfrac{2sin^2 (x) +cos (2x)+1}{2} &= 1 \\
2sin^2 (x) +cos (2x)+1 &= 2 \\
end{aligned}$$
Therefore we have derived a new representation,
$$boxed{2sin^2 (x) +cos (2x)+1 = 2}$$
This can also be written as $2 sin^2 (x) = 1 – cos (2x)$ or
$sin^2 (x) = dfrac{1 -cos (2x)}{2}$
Substituting the value of $cos^2 (x)$ in the Pythagorean identity, we get
$$begin{aligned}
dfrac{1 -cos (2x)}{2} + cos^2 (x) &= 1 \\
dfrac{1 – cos (2x)+ 2cos^2 (x) }{2} &= 1 \\
1 – cos (2x)+ 2cos^2 (x) &= 2 \\
end{aligned}$$
Therefore we have derived a new representation,
$$boxed{1 – cos (2x)+ 2cos^2 (x) = 2}$$
This can be written as $cos^2 (x) = cos (2x)+ sin^2 (x)$
Substituting this in the Pythagorean Identity we get,
$$begin{aligned}
sin^2 (x) + cos (2x)+ sin^2 (x)& = 1 \
2sin^2 (x) + cos (2x) &= 1
end{aligned}$$
Therefore we have derived a new representation,
$$boxed{2sin^2 (x) + cos (2x) = 1}$$
This can be written as $sin^2 (x) = cos^2 (x) – cos (2x)$
Substituting this in the Pythagorean Identity we get,
$$begin{aligned}
cos^2 (x) – cos (2x)+ cos^2 (x)& = 1 \
2cos^2 (x) – cos (2x) &= 1
end{aligned}$$
Therefore we have derived a new representation,
$$boxed{2cos^2 (x) – cos (2x) = 1}$$
$$a^2 – b^2 = (a-b)(a+b)$$
The Pythagorean identity can also be written as
$$begin{aligned}
sin^2 (x) &=1 – cos^2 (x) \
&=(1)^2 – (cos (x))^2 \
&=(1-cos x)(1+ cos x) \
end{aligned}$$
Therefore we have derived a new representation,
$$boxed{sin^2 (x)= (1-cos x)(1+ cos x)}$$
$$a^2 – b^2 = (a-b)(a+b)$$
The Pythagorean identity can also be written as
$$begin{aligned}
cos^2 (x) &=1 – sin^2 (x) \
&=(1)^2 – (sin (x))^2 \
&=(1-sin x)(1+ sin x) \
end{aligned}$$
Therefore we have derived a new representation,
$$boxed{cos^2 (x)= (1-sin x)(1+ sin x)}$$
$$a^2 – b^2 = (a-b)(a+b)$$
The Pythagorean identity can also be written as
$$begin{aligned}
cos^2 (x) &=1 – sin^2 (x) \
&=1 – dfrac{1}{csc^2 (x)} \\
&=dfrac{csc^2 (x) – 1}{csc^2 (x)} \\
&=dfrac{(csc x – 1)(csc x +1)}{csc^2 (x)}
end{aligned}$$
Therefore we have derived a new representation,
$$boxed{cos^2 (x)= dfrac{(csc x – 1)(csc x +1)}{csc^2 (x)}}$$
$$a^2 – b^2 = (a-b)(a+b)$$
The Pythagorean identity can also be written as
$$begin{aligned}
sin^2 (x) &=1 – cos^2 (x) \
&=1 – dfrac{1}{sec^2 (x)} \\
&=dfrac{sec^2 (x) – 1}{sec^2 (x)} \\
&=dfrac{(sec x – 1)(sec x +1)}{sec^2 (x)}
end{aligned}$$
Therefore we have derived a new representation,
$$boxed{cos^2 (x)= dfrac{(sec x – 1)(sec x +1)}{sec^2 (x)}}$$
A_{triangle ABC}=24
$$
$24=dfrac{ABcdot 8}{2}$
$24=4AB$
$AB=dfrac{24}{4}$
$AB=6$ inches
$AC^2=6^2+8^2$
$AC^2=36+64$
$AC^2=100$
$AC=sqrt{100}$
$AC=10$ inches
$$
color{#c34632}{cos^2 theta + sin^2 theta = 1}
$$
we obtain
$$
cos^2 ( theta – pi ) + sin^2 ( theta – pi ) =1
$$
$$
color{#c34632}{cos (2 theta) = cos^2 theta – sin^2 theta}
$$
we obtain
$$
begin{align*}
cos^2 (2 omega) – sin^2 (2 omega) &= cos ( 2 cdot 2 omega)\
&= cos ( 4 omega)
end{align*}
$$
$$
color{#c34632}{tan x = frac {sin x}{cos x}}
$$
we obtain
$$
frac {sin theta}{cos theta} = tan theta
$$
b) $cos^2 (2 omega) – sin^2 (2 omega) = cos ( 4 omega)$
c) $frac {sin theta}{cos theta} = tan theta$
cos^2(theta-pi)+sin^2(theta-pi)
$$
cos^2(theta-pi)+sin^2(theta-pi)=1
$$
$cos^2 x+sin^2 x=1$.
cos^2 (2w)-sin^2 (2w)
$$
cos^2 (2w)-sin^2 (2w)=cos (2cdot 2w)=cos(4w)
$$
$cos 2x=cos^2 x-sin^2 x$.
dfrac{sintheta}{costheta}
$$
dfrac{sintheta}{costheta}=tantheta
$$
$tan x=dfrac{sin x}{cos x}$.
b) $cos(4w)$;
c) $tan theta$
Using the formulas
$$
begin{align*}
color{#c34632}{a^2 – b^2 = (a-b)(a+b)}\
color{#c34632}{(a + b)^2 = a^2 + 2ab + b^2}
end{align*}
$$
Now
$$
frac {x^2 -4}{x^2 +4x+4} = frac {(x-2)(x+2)}{(x+2)^2} = frac {x-2}{x+2}
$$
Therefore
$$
frac {x^2 -4}{x^2 +4x+4} = frac {x-2}{x+2}
$$
$$
begin{align*}
frac {2x^2 – 5x-3}{4x^2 + 4x +1} &= frac {2x^2-6x+x-3}{(2x+1)^2}\
&= frac {2x(x-3)+(x-3)}{(2x+1)^2}\
&= frac {(2x+1)(x-3)}{(2x+1)^2}\
&= frac {x-3}{2x+1}
end{align*}
$$
Therefore
$$
frac {2x^2 – 5x-3}{4x^2 + 4x +1} = frac {x-3}{2x+1}
$$
b) $frac {2x^2 – 5x-3}{4x^2 + 4x +1} = frac {x-3}{2x+1}$
$y=ax^2+bx+c$
$0=a(0)^2+b(0)+c$ (Substituting with the point $(0, 0)$)
$c=0$ (1)
$9=a(3)^2+b(3)+c$ (Substituting with the point $(3, 9)$)
$9=9a+3b+c$ (2)
$0=a(6)^2+b(6)+c$ (Substituting with the point $(6, 0)$)
$0=36a+6b+c$ (3)
Eliminating $c$ from equations
$c=0$ (1)
Substituting 0 for $c$ in equations (2) and (3)
$9=9a+3b+0$
$3=3a+b$ (2)
$0=36a+6b+0$
$0=6a+b$ (3)
Subtracting (2) from (3)
$0=6a+b$ (3)
–
$3=3a+b$ (2)
—————————-
$-3=3a$
$$
a=-1
$$
Substituting in equation (2)
$3=-3+b$ (2)
$$
b=6
$$
The solution is: $a=-1$ $b=6$ $c=0$
The equation is: $y=-x^2+6x$
$0=-(0)^2+6(0)+0$ (1)
$0=0+0+0$ checkmark
$9=-(3)^2+3(6)+0$ (2)
$9=-9+18+0$ checkmark
$0=-(6)^2+6(6)+0$ (3)
$0=-36+36$ checkmark
$$
begin{align*}
frac {3x}{x+2} + frac {7}{x-2} &= 3\
frac {3x(x-2)+7 (x+2)}{(x+2)(x-2)} &= 3\
frac {3x^2 – 6x + 7x +14}{x^2 – 4} &= 3\
frac {3x^2 + x + 14}{x^2 – 4} &= 3\
3x^2 + x + 14 &= 3(x^2 -4)\
3x^2 + x + 14 &= 3x^2 -12\
x + 14 &= – 12\
x &= -26
end{align*}
$$
The solution of equation is
$$
x = -26
$$
$$
(x+2)^2 + (x-2)^2 = (2x)^2
$$
Using the formulas
$$
begin{align*}
color{#c34632}{(a+b)^2 = a^2 + 2ab + b^2}\
color{#c34632}{(a-b)^2 = a^2 – 2ab + b^2}
end{align*}
$$
Then
$$
begin{align*}
x^2 + 4x +4 + x^2 -4x + 4 &= 4x^2\
2x^2 + 8 &= 4x^2\
2x^2 &= 8\
x^2 &= 4\
x &= sqrt 4\
x &= pm 2
end{align*}
$$
The solutions of equation are
$$
x= pm 2
$$
$$
begin{align*}
3x^3 + 12x &= 21x^2 + 84\
3x^3 – 21x^2 + 12x – 84 &= 0\
3 (x^3 -7x^2 +4x -28) &= 0\
x^3 – 7x^2 + 4x -28 &= 0\
x^2 (x-7) + 4 (x-7) &= 0\
(x-7)( x^2 +4) &= 0\
x-7 = 0 qquad x^2 + 4 &= 0\
x=7 qquad x^2 &= -4\
x &= 2_i
end{align*}
$$
The solution of equation is
$$
x=7
$$
$$
10 ln 4 + ln x = 16
$$
According to the formulas
$$
begin{align*}
color{#c34632}{a ln b = ln b^a}\
color{#c34632}{ln a + ln b = ln ab}
end{align*}
$$
we obtain
$$
begin{align*}
ln 4^{10} + ln x &= 16\
ln 4^{10} x &= 16\
4^{10} x &= e^{16}\
x &= 4^{-10} e^{16}
end{align*}
$$
The solution of equation is
$$
x = 4^{-10} e^{16}
$$
b) $x=pm 2$
c) $x=7$
d) $x = 4^{-10} e^{16}$
Domain: $(-9,infty)$
(-infty,infty)
$$
$log_5(x+9)=5$
$x+9=5^5$
$x=3125-9$
$$
x=3116
$$
x=0Rightarrow y=log_5 (0+9)-5=log_5 9-5approx -3.6
$$
$x=log_5(y+9)-5$
$x+5=log_5(y+9)$
$y+9=5^{x+5}$
$y=5^5cdot 5^x-9$
$y=3125(5^x)-9$
$f^{-1}(x)=3125(5^x)-9$
(-infty,infty)
$$
(-9,infty)
$$
{color{#4257b2}text{a)}}
$$
Solution to this example is given below
Manipulating left side
$$
begin{align*}
&cot left(xright)+tan left(xright)&&boxed{text{Simplify}}\
&frac{cos left(xright)}{sin left(xright)}+tan left(xright)&&boxed{text{Using the Basic Trigonometric identity }}\
&frac{cos left(xright)}{sin left(xright)}+frac{sin left(xright)}{cos left(xright)}&&boxed{text{Using the Basic Trigonometric identity}}\
&frac{cos ^2left(xright)}{sin left(xright)cos left(xright)}+frac{sin ^2left(xright)}{sin left(xright)cos left(xright)}&&boxed{text{Adjust fractions based on the LCM}}\
&frac{cos ^2left(xright)+sin ^2left(xright)}{cos left(xright)sin left(xright)}&&boxed{text{Add fractions}}\
&frac{1}{cos left(xright)sin left(xright)}&&boxed{text{Use the following identity}}\
end{align*}
$$
Manipulating right side
$$
begin{align*}
&sec left(xright)csc left(xright)&&boxed{text{Simplify}}\
&frac{1}{sin left(xright)}sec left(xright)&&boxed{text{Using the Basic Trigonometric identity}}\
&frac{1}{cos left(xright)}cdot frac{1}{sin left(xright)}&&boxed{text{Using the Basic Trigonometric identity}}\
&frac{1}{cos left(xright)sin left(xright)}&&boxed{text{Simplify}}\\
end{align*}
$$
We showed hat the two sides
could take the same form
$$
begin{align*}
&boxed{{color{#c34632}text{True}} }&&boxed{text{Final solution}}\
end{align*}
$$
{color{#4257b2}text{b)}}
$$
Solution to this example is given below
Manipulating right side
$$
begin{align*}
&frac{1}{1+tan ^2left(xright)}&&boxed{text{Simplify}}\
&frac{1}{left(frac{sin left(xright)}{cos left(xright)}right)^2+1}&&boxed{text{Using the Basic Trigonometric identity }}\
&frac{1}{frac{sin ^2left(xright)}{cos ^2left(xright)}+1}&&boxed{text{Raise the denominator to the second power}}\
&frac{1}{frac{sin ^2left(xright)+cos ^2left(xright)}{cos ^2left(xright)}}&&boxed{text{Simplify}}\
&frac{cos ^2left(xright)}{sin ^2left(xright)+cos ^2left(xright)}&&boxed{text{Fraction rules}}\
&cos ^2left(xright)&&boxed{text{Use the following identity}}\
&left(1+sin left(xright)right)left(1-sin left(xright)right)&&boxed{text{Use the following identity}}\
end{align*}
$$
We showed hat the two sides
could take the same form
$$
begin{align*}
&boxed{{color{#c34632}text{True}} }&&boxed{text{Final solution}}\
end{align*}
$$
{color{#4257b2}text{c)}}
$$
Solution to this example is given below
Manipulating left side
$$
begin{align*}
&frac{1}{sec ^2left(xright)}+frac{1}{csc ^2left(xright)}&&boxed{text{Simplify}}\
&frac{1}{left(frac{1}{sin left(xright)}right)^2}+frac{1}{sec ^2left(xright)}&&boxed{text{Using the Basic Trigonometric identity }}\
&frac{1}{left(frac{1}{sin left(xright)}right)^2}+frac{1}{left(frac{1}{cos left(xright)}right)^2}&&boxed{text{Using the Basic Trigonometric identity}}\
&cos ^2left(xright)+sin ^2left(xright)&&boxed{text{Simplify}}\
&1&&boxed{text{Use the following identity}}\
end{align*}
$$
We showed hat the two sides
could take the same form
$$
begin{align*}
&boxed{{color{#c34632}text{True}} }&&boxed{text{Final solution}}\
end{align*}
$$
{color{#4257b2}text{d)}}
$$
Solution to this example is given below
Manipulating left side
$$
begin{align*}
&sec ^2left(xright)-csc ^2left(xright)&&boxed{text{Simplify}}\
&sec ^2left(xright)-1-cot ^2left(xright)&&boxed{text{Use the following identity }}\
&-cot ^2left(xright)+tan ^2left(xright)&&boxed{text{Use the following identity}}\
end{align*}
$$
We showed hat the two sides
could take the same form
$$
begin{align*}
&boxed{{color{#c34632}text{True}} }&&boxed{text{Final solution}}\
end{align*}
$$
{color{#4257b2}text{e)}}
$$
Solution to this example is given below
Manipulating left side
$$
begin{align*}
&frac{sec left(xright)}{tan left(xright)+cot left(xright)}&&boxed{text{Simplify}}\
&frac{frac{1}{cos left(xright)}}{cot left(xright)+tan left(xright)}&&boxed{text{Using the Basic Trigonometric identity }}\
&frac{frac{1}{cos left(xright)}}{frac{cos left(xright)}{sin left(xright)}+tan left(xright)}&&boxed{text{Using the Basic Trigonometric identity}}\
&frac{frac{1}{cos left(xright)}}{frac{cos left(xright)}{sin left(xright)}+frac{sin left(xright)}{cos left(xright)}}&&boxed{text{Using the Basic Trigonometric identity}}\
&frac{1}{cos left(xright)left(frac{cos left(xright)}{sin left(xright)}+frac{sin left(xright)}{cos left(xright)}right)}&&boxed{text{Apply the fraction rule}}\
&frac{sin left(xright)}{cos ^2left(xright)+sin ^2left(xright)}&&boxed{text{Simplify}}\
&sin left(xright)&&boxed{text{Use the following identity}}\
end{align*}
$$
We showed hat the two sides
could take the same form
$$
begin{align*}
&boxed{{color{#c34632}text{True}} }&&boxed{text{Final solution}}\
end{align*}
$$
{color{#4257b2}text{f)}}
$$
Solution to this example is given below
Manipulating left side
$$
begin{align*}
&sin ^2left(xright)cot ^2left(xright)+cos ^2left(xright)tan ^2left(xright)&&boxed{text{Simplify}}\
&cos ^2left(xright)left(frac{sin left(xright)}{cos left(xright)}right)^2+cot ^2left(xright)sin ^2left(xright)&&boxed{text{Using the Basic Trigonometric identity }}\
&left(frac{cos left(xright)}{sin left(xright)}right)^2sin ^2left(xright)+left(frac{sin left(xright)}{cos left(xright)}right)^2cos ^2left(xright)&&boxed{text{Using the Basic Trigonometric identity}}\
&cos ^2left(xright)+sin ^2left(xright)&&boxed{text{Simplify}}\
&1&&boxed{text{Use the following identity}}\
end{align*}
$$
We showed hat the two sides
could take the same form
$$
begin{align*}
&boxed{{color{#c34632}text{True}} }&&boxed{text{Final solution}}\
end{align*}
$$
{color{#4257b2}text{g)}}
$$
Solution to this example is given below
Manipulating left side
$$
begin{align*}
&frac{csc ^2left(xright)-1}{csc ^2left(xright)}&&boxed{text{Simplify}}\
&frac{-1+left(frac{1}{sin left(xright)}right)^2}{left(frac{1}{sin left(xright)}right)^2}&&boxed{text{Using the Basic Trigonometric identity }}\
&frac{left(frac{1}{sin left(xright)}right)^2-1}{left(frac{1}{sin left(xright)}right)^2}&&boxed{text{Using the Basic Trigonometric identity}}\
&1-sin ^2left(xright)&&boxed{text{Simplify}}\
&cos ^2left(xright)&&boxed{text{Use the following identity}}\
end{align*}
$$
We showed hat the two sides
could take the same form
$$
begin{align*}
&boxed{{color{#c34632}text{True}} }&&boxed{text{Final solution}}\
end{align*}
$$
{color{#4257b2}text{h)}}
$$
Solution to this example is given below
Manipulating left side
$$
begin{align*}
&frac{sin left(xright)}{cos left(xright)}+frac{cos left(xright)}{sin left(xright)}-2=0&&boxed{text{Subtract }2 text{ from both sides}}\
&frac{sin left(xright)}{cos left(xright)}+frac{cos left(xright)}{sin left(xright)}-frac{2}{1}&&boxed{text{Convert element to a fraction }2=frac{2}{1}}\
&frac{sin ^2left(xright)}{cos left(xright)sin left(xright)}+frac{cos ^2left(xright)}{cos left(xright)sin left(xright)}-frac{2cos left(xright)sin left(xright)}{cos left(xright)sin left(xright)}&&boxed{text{Adjust fraction based on the LCM}}\
&frac{sin ^2left(xright)+cos ^2left(xright)-2cos left(xright)sin left(xright)}{cos left(xright)sin left(xright)}=0&&boxed{text{Simplify}}\
&sin ^2left(xright)+cos ^2left(xright)-2cos left(xright)sin left(xright)=0&&boxed{text{Simplify}}\
&1-2cos left(xright)sin left(xright)=0&&boxed{text{Use the following identity}}\
&1-sin left(2xright)=0&&boxed{text{Use the following identitty}}\
&1-sin left(2xright)-1=0-1&&boxed{text{Subtract 1 from both sides}}\
&-sin left(2xright)=-1&&boxed{text{Simplify}}\
&frac{-sin left(2xright)}{-1}=frac{-1}{-1}&&boxed{text{Divide both sides by }-1}\
&sin left(2xright)=1&&boxed{text{Simplify}}\
end{align*}
$$
We showed hat the two sides
could take the same form
$$
begin{align*}
&boxed{{color{#c34632}text{True}} }&&boxed{text{Final solution}}\
end{align*}
$$
color{#4257b2} text{ a) True }
$$
$$
color{#4257b2} text{ b) True }
$$
$$
color{#4257b2} text{ c) True }
$$
$$
color{#4257b2} text{ d) True }
$$
$$
color{#4257b2} text{ e) True }
$$
$$
color{#4257b2} text{ f) True }
$$
$$
color{#4257b2} text{ g) True }
$$
$$
color{#4257b2} text{ h) True }
$$
a. We have the expression $sin (x) = -1$ \
We know that $sin left(dfrac{pi}{2}right) = 1$
Also, sine is negative in the third and fourth quadrant. This implies $sin (pi + theta) = – sin theta$ and $sin (2pi – theta) = – sin theta$
Taking $theta = dfrac{pi}{2}$
$$begin{aligned}
sin left(pi+dfrac{pi}{2}right) &= -sin left(dfrac{pi}{2}right) \\
sin left(dfrac{3 pi}{2}right) &= -1 \\
end{aligned}$$
and
$$begin{aligned}
sin left(2pi-dfrac{pi}{2}right) &= -sin left(dfrac{pi}{2}right) \\
sin left(dfrac{3 pi}{2}right) &= -1 \\
end{aligned}$$
Therefore $x = dfrac{3 pi}{2}$ for $sin (x) = -1$
We can rewrite it as $2 cos (x) = 1$ or $cos (x) = dfrac{1}{2}$
Now taking $cos^{-1}$ on both sides, we get
$$begin{aligned}
cos^{-1}[cos (x)] &=cos^{-1}left(dfrac{1}{2}right) \\
x&=dfrac{pi}{3}
end{aligned}$$
Now we have x$=dfrac{pi}{3}$
Since cosine is positive in the fourth quadrant,
$cos (2pi – theta) = cos (theta)$
This implies, taking $theta = dfrac{pi}{3}$
$cos (2pi – dfrac{pi}{3}) = cos left(dfrac{pi}{3}right)$
or
$cos left(dfrac{5pi}{3}right) = cos left(dfrac{pi}{3}right)$
Thus, we have x = $dfrac{pi}{3}$ and x= $dfrac{5pi}{3}$
for $2 cos (x) – 1 = 0$ where $0 le x < 2 pi$
Taking $tan^{-1}$ on both sides, we get
$$begin{aligned}
tan^{-1} [tan (x)] &= tan^{-1} (1) \\
x&=dfrac{pi}{4}
end{aligned}$$
We know $tan$ is positive in the third quadrant. Therefore, we have
$tan (pi + theta) = tan (theta)$
Putting $theta = dfrac{pi}{4}$ we get
$$begin{aligned}
tan left(pi + dfrac{pi}{4}right) &= tan left(dfrac{pi}{4}right) \\
tan left(dfrac{5pi}{4}right) &= tan left(dfrac{pi}{4}right)
end{aligned}$$
Thus, we have x = $dfrac{pi}{4}$ and x= $dfrac{5pi}{4}$
for $tan(x) = 1$ where $0 le x < 2 pi$
We will first simplify this expression \
$$begin{aligned}
2 sin (x) &= 4 sin (x) + 1 \
4 sin (x) – 2 sin (x) &= -1 \
2 sin (x) &= -1 \
sin(x)&= -dfrac{1}{2} \
end{aligned}$$
We know, $sin left(dfrac{pi}{6}right) = dfrac{1}{2}$
Also, sine is negative in the third and fourth quadrant. This implies $sin (pi + theta) = – sin theta$ and $sin (2pi – theta) = – sin theta$
Accordingly,
$$begin{aligned}
sin left(pi+dfrac{pi}{6}right) &= -sin left(dfrac{pi}{6}right) \\
sin left(dfrac{7 pi}{6}right) &= -dfrac{1}{2} \\
end{aligned}$$
and
$$begin{aligned}
sin left(2pi-dfrac{pi}{6}right) &= -sin left(dfrac{pi}{6}right) \\
sin left(dfrac{11 pi}{6}right) &= -dfrac{1}{2} \\
end{aligned}$$
Thus, we have x = $dfrac{7pi}{6}$ and x= $dfrac{11pi}{6}$
for $2 sin (x) = 4 sin (x) + 1$ where $0 le x < 2 pi$
$$
color{#c34632}{sin^2 theta + cos^2 theta = 1}
$$
Sine takes value with $y-text{axis}$
$$
sin theta = frac yr
$$
Since $r=1$ we obtain
$$
sin theta = y
$$
Cosine takes value with $x-text{axis}$
$$
cos theta = frac xr
$$
Since $r=1$ we obtain
$$
cos theta = x
$$
Now we have
$$
y^2 + x^2 =1
$$
We have $y=- frac 38$, then
$$
begin{align*}
x^2 &+ left( – frac 38 right)^2 = 1\
x^2 &+ frac {9}{64} = 1\
x^2 &= 1 – frac {9}{64}\
x^2 &= frac {64-9}{64}\
x^2 &= frac {55}{64}\
x &= pm sqrt {frac {55}{64}}\
x &= pm frac {sqrt {55}}{8}
end{align*}
$$
The point is
$$
(x,y) = left( pm frac {sqrt {55}}{8} , – frac 38 right)
$$
(x,y) = left( pm frac {sqrt {55}}{8} , – frac 38 right)
$$
Pleft(x,-dfrac{3}{8}right)
$$
$sintheta=dfrac{y}{r}=dfrac{y}{1}=y$
$x^2+left(-dfrac{3}{8}right)^2=1$
$x^2+dfrac{9}{64}=1$
$x^2=1-dfrac{9}{64}$
$x^2=dfrac{55}{64}$
$$
x=pmdfrac{sqrt{55}}{8}
$$
$sin^2theta+cos^2theta=1$.
$P_2left(dfrac{sqrt{55}}{8},-dfrac{3}{8}right)$
$P_2left(dfrac{sqrt{55}}{8},-dfrac{3}{8}right)$
Using the formula
$$
color{#c34632}{c^2 = a^2 + b^2 – 2ab cos theta}
$$
Therefore
$$
begin{align*}
10^2 &= 5^2 + 9^2 – 2 cdot 5cdot 9 cos theta\
100 &= 25 + 81 – 90 cos theta\
-6 &= -90 cos theta
end{align*}
$$
Multiply by $(-1)$
$$
begin{align*}
90cos theta &= 6\
cos theta &= frac {6}{90}\
cos theta &= 0.06
end{align*}
$$
By applying the inverse function we get
$$
begin{align*}
theta &= arccos 0.06\
theta &= 1.51\
theta &= 86.5^{text{textdegree}}
end{align*}
$$
theta = 86.5^{text{textdegree}}
$$
$BC=9$
$$
AC=10
$$
AC>BC>AB
$$
$10^2=5^2+9^2-2(5)(9)cos B$
$100=25+81-90cos B$
$100=106-90cos B$
$90cos B=106-100$
$90cos B=6$
$cos B=dfrac{6}{90}$
$cos B=dfrac{1}{15}$
$$
B=cos^{-1}dfrac{1}{15}approx 86.19text{textdegree}
$$
Bapprox 86.19text{textdegree}
$$
f(x)=2sinleft(x+dfrac{pi}{2}right)
$$
$$
T=dfrac{2pi}{1}=2pi
$$
We start with the parent function $f_0(x)=sin x$. We shift $f_0$ $dfrac{pi}{2}$ units to the left to get $f_1(x)=sinleft(x+dfrac{pi}{4}right)$,then we vertically stretch $f_1$ by a factor of 2 to get $f(x)$:
y=cos(2x)-1
$$
$$
T=dfrac{2pi}{2}=pi
$$
We start with the parent function $f_0(x)=cos x$. We horizontally shrink $f_0$ by a factor of 2 to get $f_1(x)=cos (2x)$, then we shift $f_1$ 1 unit down to get $f(x)$:
y=tanleft(x+dfrac{pi}{4}right)
$$
T=dfrac{pi}{1}=pi
$$
We determine the period $T$ of the function:
We start with the parent function $f_0(x)=tan x$. We shift $f_0$ $dfrac{pi}{2}$ units to the left to get $f(x)$:
f(x)=sin x+1
$$
f(x)=cos (2x)
$$
$$
5=c
$$
$$
1=4a+2b+c
$$
$$
17=36a+6b+c
$$
Replace $c$ with its known values in the other equations:
$$
5=c
$$
$$
1=4a+2b+5
$$
$$
17=36a+6b+5
$$
Rewrite the equations:
$$
c=5
$$
$$
-2=2a+b
$$
$$
2=6a+b
$$
Subtract the third equation from the second equation:
$$
c=5
$$
$$
-4=-4a
$$
$$
2=6a+b
$$
Solve the second equation to $a$:
$$
c=5
$$
$$
1=a
$$
$$
2-6a=b
$$
Determine $b$:
$$
c=5
$$
$$
a=1
$$
$$
b=2-6a=2-6(1)=-4
$$
Thus the equation then becomes $f(x)=x^2-4x+5$
$$
x=dfrac{-b}{2a}=dfrac{4}{2}=2
$$
The vertex is the intersection of the axis of symmetry and the function:
$$
f(2)=2^2-4(2)+5=4-8+5=1
$$
Thus the vertex is $(2,1)$.
b. $(2,1)$
$$
y= sqrt[3]{x-4} + 2
$$
Now, the locator point is
$$
(4,2)
$$
(4,2)
$$
f(x)=sqrt[3]{x-4}+2
$$
$$
0.52, 0.55, 0.58, …, 2.02
$$
where
$$
a= 0.52 qquad text{and} qquad d=0.03
$$
a) To calculate the number of terms, we need to subtract the first term $0.52$ from the last term $2.02$
$$
2.02 – 0.52 = 1.5
$$
Since
$$
frac {1.5}{d} = frac {1.5}{0.03} = 50
$$
we have $50$ terms.
Therefore
$$
n=50
$$
$$
color{#c34632}{ S = frac n2 [ 2a + (n-1) d]}
$$
we obtain
$$
begin{align*}
S &= frac {50}{2} [ 2 cdot 0.52 + (50-1) 0.03 ]\
S &= 25 [ 1.04 + 49 cdot 0.03 ]\
S &= 25 [ 1.04 + 1.47 ]\
S &= 25 cdot 2.51\
S &= 62.75
end{align*}
$$
The sum is
$$
S= 62.75
$$
b) $S= 62.75$
a) According to the formula
$$
color{#c34632}{ sin^2 theta + cos^2 theta = 2}
$$
we obtain
$$
begin{align*}
( sin theta + cos theta )^2 &= 1 + 2 sin theta cos theta\
sin^2 theta + 2 sin theta cos theta + cos^2 theta &= 1 + 2 sin theta cos theta\
sin^2 theta + cos^2 theta + 2 sin theta cos theta &= 1 + 2 sin theta cos theta\
1 + 2 sin theta cos theta &= 1 + 2 sin theta cos theta
end{align*}
$$
The equation is identity.
$$
color{#c34632}{tan theta = frac {sin theta}{cos theta}}
$$
$$
color{#c34632}{cot theta = frac {cos theta}{sin theta}}
$$
$$
color{#c34632}{sec theta = frac {1}{cos theta}}
$$
$$
color{#c34632}{csc theta = frac {1}{sin theta}}
$$
we obtain
$$
begin{align*}
tan theta + cot theta &= sec theta csc theta\
frac {sin theta}{cos theta} + frac {cos theta}{sin theta} &= frac {1}{cos theta} cdot frac {1}{sin theta}\
frac {sin^2 theta + cos^2 theta}{cos theta sin theta} &= frac {1}{cos theta sin theta}\
frac {1}{cos theta sin theta} &= frac {1}{cos theta sin theta}
end{align*}
$$
The equation is identity.
$$
color{#c34632}{tan theta = frac {sin theta}{cos theta}}
$$
$$
color{#c34632}{sec theta = frac {1}{cos theta}}
$$
we obtain
$$
begin{align*}
( tan theta cos theta ) left( sin^2 theta + frac {1}{sec^2 theta} right) &= sin theta\
left( frac {sin theta}{cos theta} cdot cos theta right) left( sin^2 theta + frac {1}{frac {1}{cos^2 theta}} right) &= sin theta\
sin theta ( sin^2 theta + cos^2 theta ) &= sin theta\
sin theta &= sin theta
end{align*}
$$
The equation is identity.
b) $frac {1}{cos theta sin theta} = frac {1}{cos theta sin theta}$
c) $sin theta = sin theta$
$$
begin{align*}
frac {2 + sin^2 theta}{3} &= frac 34\
2 + sin^2 theta &= frac 34 cdot 3\
2 + sin^2 theta &= frac 94\
sin^2 theta &= frac 94 – 2\
sin^2 theta &= frac 14\
sin theta &= pm sqrt {frac 14}\
sin theta &= pm frac 12
end{align*}
$$
By applying the inverse function we get
$$
theta = arcsin left( pm frac 12 right)
$$
solution:
$$
begin{align*}
theta &= arcsin frac 12\
theta &= frac {pi}{6} qquad theta = frac {5 pi}{6}
end{align*}
$$
solution:
$$
begin{align*}
theta &= arcsin left( – frac 12 right)\
theta &= frac {7pi}{6} qquad theta = frac {11 pi}{6}
end{align*}
$$
The solutions of the equation are
$$
begin{align*}
theta &= frac {pi}{6} qquad theta = frac {5 pi}{6}\
theta &= frac {7pi}{6} qquad theta = frac {11 pi}{6}
end{align*}
$$
begin{align*}
theta &= frac {pi}{6} qquad theta = frac {5 pi}{6}\
theta &= frac {7pi}{6} qquad theta = frac {11 pi}{6}
end{align*}
$$
dfrac{2+sin^2theta}{3}=dfrac{3}{4},0leqtheta<2pi
$$
$8+4sin^2theta=9$
$sintheta=pmdfrac{1}{2}$
$theta_1=dfrac{pi}{6}$
$theta_2=dfrac{5pi}{6}$
$sintheta=-dfrac{1}{2}$
$theta_3=dfrac{7pi}{6}$
$theta_4=dfrac{11pi}{6}$
left{dfrac{pi}{6},dfrac{5pi}{6},dfrac{7pi}{6},dfrac{11pi}{6}right}
$$
$$
3(1)+(2)-(-1)=3+2+1=6
$$
b. Yes
$$cos (x) = dfrac{1}{2} rightarrow(1)$$
Solving ,
$$begin{aligned}
cos (x)&= cos 30^{o}\
&boxed{x=30}\
end{aligned}$$
Sketching the graph of equation (1) :
$$
x^2 – 3x – 4 =0
$$
The solution of a square equation $a^2 + bx + c =0$ is shape
$$
x= frac {-b pm sqrt {b^2 -4ac}}{2a}
$$
Now, we have
$$
begin{align*}
x &= frac {3 pm sqrt {9+16}}{2}\
x &= frac {3 pm sqrt {25}}{2}\
x &= frac {3 pm 5}{2}\
x &= 4 qquad x = -1
end{align*}
$$
$$
x^2 – 3x – 4 geq 0
$$
Discriminant is
$$
begin{align*}
D &= b^2 – 4ac = (-3)^2 – 4 cdot 1 cdot (-4)= 9 + 16 = 25 > 0
end{align*}
$$
then
$$
x^2 – 3x – 4 geq 0
$$
for
$$
xin ( – infty , -1 ] cup [ 4, + infty )
$$
$$
x^2 -3x -4 leq 0
$$
The solution of the inequality is
$$
x in [-1, 4]
$$
b)$xin ( – infty , -1 ] cup [ 4, + infty )$
c)$x in [-1, 4]$
$$
color{#c34632}{ d = sqrt {(x_2 – x_1)^2 + ( y_2 – y_1)^2}}
$$
a) Let $(x_1, y_1)=(x,y)$ and $(x_2, y_2)=(-3, y)$
Now we have
$$
begin{align*}
d &= sqrt {(-3 – x)^2 + ( y – y)^2}\
d &= sqrt {(-3-x) +0}\
d &= sqrt {(-3-x)^2}\
d &= |-3-x|\
d &= |3+x|
end{align*}
$$
Now we have
$$
d = sqrt {(-3-x)^2 + (2-y)^2}
$$
b)$d = sqrt {(-3-x)^2 + (2-y)^2}$
$$
(-3,y)
$$
$d=sqrt{(x+3)^2}$
$$
d=|x+3|
$$
$d=sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.
$$
(-3,2)
$$
$d=sqrt{(x+3)^2+(y-2)^2}$
b) $d=sqrt{(x+3)^2+(y-2)^2}$
Since we know that the external angle is equal to the sum of interior opposite angles. Thus, $angle POQ$ can be expressed in terms of $alpha$ and $beta$ as follows :
$$m (angle POQ ) = alpha – beta rightarrow (1)$$
Method 1 $rightarrow$ “Distance Formula”
Method 1 $rightarrow$ “Law of cosines”
Coordinate of $P rightarrow (cos alpha, sin alpha)$.
Coordinate of $Q rightarrow (cos beta, sin beta)$
Using Pythagorean indentity and substitute the value of the coordinate, we get
$$begin{aligned}
d&= sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\
PQ&= sqrt{(cos beta – cos alpha)^2+ (sin beta – sin alpha)^2}\
PQ&= sqrt{cos^2beta + cos^2 beta – 2 cosalpha cos beta + sin^2 beta + sin^2alpha- 2sinalphasin beta}\
&boxed{PQ=sqrt{2 – 2sin alpha sin beta – 2 cos alpha cos beta}} rightarrow (sin^2 x + cos^2 = 1)\
end{aligned}$$
Using law of cosines to write an expression for the length of side $PQ$.
Since it is an unit circle , the radius $OP$ & $OQ$ will be of one unit.
$$begin{aligned}
PQ^2&= OP^2 + OQ^2-2(OP)(OQ)cos(angle POQ)\
text{Substituting the values, }\
PQ^2&= (1)^2 + (1)^2 – 2(1)(1) cos (alpha – beta) rightarrow(text{ From} (1))\
PQ^2&= 2 – 2 cos (alpha – beta)
text{Taking square root both sides,}\
PQ&= pmsqrt{2 – 2cos (alpha – beta) }
end{aligned}$$
Using the result of the item $d$ and $e$ :
$$2 – 2 sin alpha sin beta – 2 cos alpha cos beta = 2 – 2 cos (alpha – beta)$$
$$cos (alpha – beta) = cos alpha cos beta + sin alpha sin beta$$
$$cos (alpha – beta) = cos alpha cos beta + sin alpha sin beta$$
Danny and Damian want to know the formula of $cos (alpha + beta)$ Damian has proposed to rewrite the above expression as
$$cos [alpha – (-beta)]$$
Accordingly,
$$cos [alpha – (-beta)]= cos alpha cos (-beta) + sin alpha sin (-beta)$$
Now, sine is an odd function and cosine an even one.
This implies
$sin (-x) = -sin x$ and $cos (-x) = cos x$
Therefore in our equation above $sin (-beta) = -sin beta$ and $cos (-beta) = cos beta$
Therefore,
$$begin{aligned}
cos [alpha – (-beta)]&= cos alpha cos (-beta) + sin alpha sin (-beta) \
&= cos alpha cos (beta) + sin alpha [-sin (beta)] \
&=cos alpha cos beta – sin alpha sin beta \
end{aligned}$$
Thus we have
$$boxed{cos (alpha + beta) = cos alpha cos beta – sin alpha sin beta}$$
$$cos left(dfrac{pi}{2} -thetaright) = sin (theta)$$
and
$$sin left(dfrac{pi}{2} -thetaright) = cos (theta)$$
Now we know,
$sin (theta) = cos left(dfrac{pi}{2} -thetaright)$ and
$cos (alpha – beta) = cos alpha cos beta + sin alpha sin beta$
Therefore,
$$begin{aligned}
sin (alpha + beta) &= cos left(dfrac{pi}{2} -[alpha + beta]right) \\
&=cos left(dfrac{pi}{2} – alpha – betaright) \\
&=cos left([dfrac{pi}{2} – alpha] – betaright) \\
&=cos left(dfrac{pi}{2} -alpharight) cos beta + sin left(dfrac{pi}{2} -alpharight) sin beta \\
&=sin alpha cos beta + cos alpha sin beta \
end{aligned}$$
Therefore, we have the Angle Sum Identity for sine,
$$boxed{sin (alpha + beta)=sin alpha cos beta + cos alpha sin beta}$$
$$sin (alpha + beta) = sin alpha cos beta + cos alpha sin beta$$
Danny and Damian want to know the formula of $sin (alpha – beta)$
We can rewrite the above expression as
$$sin [alpha + (-beta)]$$
Accordingly,
$$sin [alpha + (-beta)]= sin alpha cos (-beta) + cos alpha sin (-beta)$$
Now, sine is an odd function and cosine an even one.
This implies
$sin (-x) = -sin x$ and $cos (-x) = cos x$
Therefore in our equation above $cos (-beta) = cos beta$ and $sin (-beta) = -sin beta$
Therefore,
$$begin{aligned}
sin [alpha + (-beta)]&= sin alpha cos (-beta) + cos alpha sin (-beta) \
&= sin alpha cos (beta) + cos alpha [-sin (beta)] \
&=sin alpha cos beta – cos alpha sin beta \
end{aligned}$$
Thus we have the Angle Difference identity,
$$boxed{sin (alpha – beta) = sin alpha cos beta – cos alpha sin beta}$$
First we solve $textbf{sine }$ of $boldsymbol{frac{pi}{12}}$
$$
begin{align*}
&sin left(frac{frac{pi }{6}}{2}right)&&boxed{text{Simplify}}\
&sqrt{frac{1-cos left(frac{pi }{6}right)}{2}}&&boxed{text{Using the half angle identity}}\
&sqrt{frac{1-frac{sqrt{3}}{2}}{2}}&&boxed{text{Use the following trivial identity}}\
&frac{sqrt{2-sqrt{3}}}{sqrt{4}}&&boxed{text{Apply radical rule}}\
&frac{sqrt{2-sqrt{3}}}{sqrt{2^2}}&&boxed{text{Factor the number: }4=2^2}\
&frac{sqrt{2-sqrt{3}}}{2}&&boxed{text{Simplify}}\
end{align*}
$$
$$
begin{align*}
&cos left(frac{frac{pi }{6}}{2}right)&&boxed{text{Simplify}}\
&sqrt{frac{1+cos left(frac{pi }{6}right)}{2}}&&boxed{text{Using the half angle identity}}\
&sqrt{frac{1+frac{sqrt{3}}{2}}{2}}&&boxed{text{Use the following trivial identity}}\
&frac{sqrt{2+sqrt{3}}}{sqrt{4}}&&boxed{text{Apply radical rule}}\
&frac{sqrt{2+sqrt{3}}}{sqrt{2^2}}&&boxed{text{Factor the number: }4=2^2}\
&frac{sqrt{2+sqrt{3}}}{2}&&boxed{text{Simplify}}\\
&boxed{{color{#c34632}sinfrac{pi}{12}=frac{sqrt{2-sqrt{3}}}{2}, cosfrac{pi}{12}=frac{sqrt{2+sqrt{3}}}{2}} }&&boxed{text{Final solution}}\
end{align*}
$$
$$
boxed{ color{#c34632} text{ } mathrm{Using:the:half:angle:identity}:quad sin left(frac{x}{2}right)=sqrt{frac{1-cos left(xright)}{2}}}
$$
$$
boxed{ color{#c34632} text{ }mathrm{Apply:radical:rule:}sqrt[n]{frac{a}{b}}=frac{sqrt[n]{a}}{sqrt[n]{b}},:quad mathrm{:assuming:}age :0,:bge :0}
$$
$$
boxed{ color{#c34632} text{ }mathrm{Apply:the:fraction:rule}:quad frac{frac{b}{c}}{a}=frac{b}{c:cdot :a} }
$$
$$
boxed{ color{#c34632} text{ }mathrm{Apply:radical:rule}:quad sqrt[n]{a^n}=a }
$$
color{#4257b2} text{ }sinfrac{pi}{12}=frac{sqrt{2-sqrt{3}}}{2}, cosfrac{pi}{12}=frac{sqrt{2+sqrt{3}}}{2}
$$
$$
begin{align*}
color{#c34632}{sin (alpha + beta) = sin alpha cos beta + cos alpha sin beta} tag{$star$}
end{align*}
$$
Now, we can write
$$
begin{align*}
sin frac {7 pi}{12} &= sin left( frac {4 pi}{12} + frac {3 pi}{12} right)\
&= sin left( frac {pi}{3} + frac {pi}{4} right)\
&overset{(star)}= sin frac {pi}{3} cos frac {pi}{4} + cos frac {pi}{3} sin frac {pi}{4}\
&= frac {sqrt 3}{2} cdot frac {sqrt 2}{2} + frac 12 frac {sqrt 2}{2}\
&= frac {sqrt 6}{4} + frac {sqrt 2}{4}\
&= frac {sqrt 6 + sqrt 2}{4}
end{align*}
$$
$$
color{#c34632}{ cos ( alpha + beta) = cos alpha cos beta – sin alpha sin beta}
$$
Now, we can write
$$
begin{align*}
cos left( frac {11 pi}{12} right) &= cos left( frac {8 pi}{12} + frac {3 pi}{12} right)\
&= cos left( frac {2 pi}{3} + frac {pi}{4} right)\
&= cos frac {2 pi}{3} cos frac{pi}{4} – sin frac {2 pi}{3} sin frac {pi}{4}\
&= – frac 12 cdot frac {sqrt 2}{2} – frac {sqrt 3}{2} cdot frac {sqrt 2}{2}\
&= – frac {sqrt 2}{4} – frac {sqrt 6}{4}\
&= frac {- sqrt 2 – sqrt 6}{4}
end{align*}
$$
b) $cos left( frac {11 pi}{12} right) = frac {- sqrt 2 – sqrt 6}{4}$
sindfrac{7pi}{12}
$$
$$
=sinleft(dfrac{pi}{4}+dfrac{pi}{3}right)
$$
$=sindfrac{pi}{4}cos dfrac{pi}{3}+cos dfrac{pi}{4}sin dfrac{pi}{3}$
$=dfrac{sqrt 2}{2}cdot dfrac{1}{2}+dfrac{sqrt 2}{2}cdot dfrac{sqrt 3}{2}$
$$
=dfrac{sqrt 2+sqrt 6}{4}
$$
$sin(x+y)=sin xcos y+cos xsin y$.
cosdfrac{11pi}{12}
$$
$$
=cosleft(dfrac{pi}{4}+dfrac{2pi}{3}right)
$$
$=cosdfrac{pi}{4}cos dfrac{2pi}{3}-sin dfrac{pi}{4}sin dfrac{2pi}{3}$
$=dfrac{sqrt 2}{2}left(-dfrac{1}{2}right)-dfrac{sqrt 2}{2}cdot dfrac{sqrt 3}{2}$
$$
=-dfrac{sqrt 2+sqrt 6}{4}
$$
$cos(x+y)=cos xcos y-sin xsin y$.
b) $-dfrac{sqrt 2+sqrt 6}{4}$
$$cos (alpha + beta) = cos alpha cos beta – sin alpha sin beta$$
and
$$sin (alpha + beta) = sin alpha cos beta + cos alpha sin beta$$
We can write
$$sin(2x) = sin(x+x)$$
and
$$cos(2x) =cos(x+x)$$
$$begin{aligned}
sin(x+x)&=sin x cos x + cos x sin x\
&=sin x cos x + sin x cos x \
&=2sin x cos x \
end{aligned}$$
Therefore, we have the identity,
$$boxed{sin(x+x)=2sin x cos x }$$
$$begin{aligned}
cos(x+x)&=cos x cos x sin x sin x\
&=cos^2 x – sin^2 x \
end{aligned}$$
Therefore, we have the identity,
$$boxed{cos(x+x)=cos^2 x – sin^2 x}$$
We also know the Pythagorean identity i.e.
$$cos^2 x + sin^2 x=1$$
This can be rewritten as,
$$cos^2 x =1- sin^2 x$$
Substituting this in the identity of $cos (2x)$
$$begin{aligned}
cos(2x)&= cos^2 x – sin^2 \
&=1- sin^2 x- sin^2 x \
&=1- 2sin^2 x
end{aligned}$$
Therefore, we have the identity,
$$boxed{cos(x+x)=1 – 2sin^2 x}$$
$$ sin^2 x =1 – cos^2 x $$
Substituting this in the identity of $cos (2x)$
$$begin{aligned}
cos(2x)&= cos^2 x – (1 – cos^2 x) \
&=cos^2 x – 1+ cos^2 x) \
&=2cos^2 x – 1 \
end{aligned}$$
Therefore, we have the identity,
$$boxed{cos(x+x)=2cos^2 x – 1}$$
$$f(x)=cosleft(x+dfrac{pi}{2}right)$$
We can graph the function as follows:
$$cosleft(theta+dfrac{pi}{2}right) = -sin (theta)$$
Therefore,
$$cosleft(x+dfrac{pi}{2}right) = – sin (x)$$
$$cos (alpha + beta) = cos alpha cos beta – sin alpha sin beta$$
It must be noted
$cosleft(dfrac{pi}{2}right) = 0$ and $sinleft(dfrac{pi}{2}right) = 1$
Now solving our function as per the Angle Sum Identity of cosine
$$begin{aligned}
cosleft(x+dfrac{pi}{2}right) &= cos (x) cosleft(dfrac{pi}{2}right) – sin (x) sinleft(dfrac{pi}{2}right) \\
&=cos x times 0 – sin (x) times 1 \
&= – sin(x)
end{aligned}$$
Therefore we have
$$cosleft(x+dfrac{pi}{2}right) = – sin (x)$$
We will graph $- sin(x)$
$$dfrac{cos(x)}{1 – tan(x)} – dfrac{sin(x)}{cot (x) -1} $$
We know
$tan (theta) = dfrac{sin (theta)}{cos (theta)}$
and
$cot (theta) = dfrac{cos (theta)}{sin (theta)}$
The expression has two parts. We will solve these parts separately.
$$dfrac{cos(x)}{1 – tan(x)}$$
Simplifying it we get
$$begin{aligned}
dfrac{cos(x)}{1 – tan(x)} &= dfrac{cos(x)}{1 – frac{sin (x)}{cos (x)}} \\
&= dfrac{cos(x)}{frac{cos (x)-sin (x)}{cos (x)}} \\
&= dfrac{cos(x) times cos (x)}{cos (x)-sin (x)} \\
&= dfrac{cos^2(x) }{cos (x)-sin (x)} \
end{aligned}$$
Therefore we have,
$$dfrac{cos(x)}{1 – tan(x)}=dfrac{cos^2(x) }{cos (x)-sin (x)}$$
$$dfrac{sin(x)}{cot(x)-1}$$
Simplifying it we get
$$begin{aligned}
dfrac{sin(x)}{cot(x)-1}&= dfrac{sin(x)}{frac{cos (x)}{sin (x)}-1} \\
&= dfrac{sin(x)}{frac{cos (x)-sin (x)}{sin (x)}} \\
&= dfrac{sin(x) times sin (x)}{cos (x)-sin (x)} \\
&= dfrac{sin^2(x) }{cos (x)-sin (x)} \
end{aligned}$$
Therefore we have,
$$dfrac{sin(x)}{cot(x)-1}=dfrac{sin^2(x) }{cos (x)-sin (x)}$$
$$dfrac{cos^2(x) }{cos (x)-sin (x)} – dfrac{sin^2(x) }{cos (x)-sin (x)}$$
We must note that $a^2 – b^2 = (a-b)(a+b)$
Simplifying our equation we get
$$begin{aligned}
dfrac{cos^2(x) }{cos (x)-sin (x)} – dfrac{sin^2(x) }{cos (x)-sin (x)} &= dfrac{cos^2(x)-sin^2(x) }{cos (x)-sin (x)} \\
&=dfrac{[cos(x)-sin(x)][cos(x)+sin(x)] }{cos (x)-sin (x)} \\
&=cos(x)+sin(x)\\
end{aligned}$$
Therefore we have
$$boxed{dfrac{cos(x)}{1 – tan(x)} – dfrac{sin(x)}{cot (x) -1}=cos(x)+sin(x)}$$
begin{cases}
-4x=z-2y+12\
y+z=12-x\
8x-3y+4z=1
end{cases}
$$
begin{cases}
-4x+2y-z=12\
x+y+z=12\
8x-3y+4z=1
end{cases}
$$
begin{cases}
-4x+2y-z+x+y+z=12+12\
-16x+8y-4z+8x-3y+4z=48+1
end{cases}
$$
$$
begin{cases}
-3x+3y=24\
-8x+5y=49
end{cases}
$$
$$
begin{cases}
-x+y=8\
-8x+5y=49
end{cases}
$$
$-3x=9$
$$
textcolor{#4257b2}{x=-3}
$$
$-(-3)+y=8$
$3+y=8$
$y=8-3$
$$
textcolor{#4257b2}{y=5}
$$
$5+z=12-(-3)$
$z=15-5$
$$
textcolor{#4257b2}{z=10}
$$
$y=5$
$$
z=10
$$
begin{cases}
3x+y-2z=6\
x+2y+z=7\
6x+2y-4z=12
end{cases}
$$
begin{cases}
3x+y-2z=6\
x+2y+z=7\
-3x-y+2z=-6
end{cases}
$$
$$
0=0
$$
begin{cases}
3x+y=2k+6\
x+2y=7-k
end{cases}
$$
$$
z=k
$$
$-5y=5k-15$
$y=3-k$
$$
x=7-k-2(3-k)=7-k-6+2k=1+k
$$
$y=3-k$
$z=k, k$ real number
Red: 240
P(black)=dfrac{60}{60+240}=dfrac{60}{300}=dfrac{1}{5}
$$
Red: $240$
P(black)=dfrac{60+60}{60+60+240}=dfrac{120}{360}=dfrac{1}{3}
$$
$dfrac{x+60}{x+300}=2cdotdfrac{1}{5}$
$dfrac{x+60}{x+300}=dfrac{2}{5}$
$5(x+60)=2(x+300)$
$5x+300=2x+600$
$5x-2x=600-300$
$3x=300$
$x=dfrac{300}{3}$
$$
x=100
$$
$x$=the number of black jellybeans that must be added so that the probability to draw a black one is doubled.
We determine the probability to draw a black jellybean:
b) $dfrac{1}{3}$
c) $100$;
d) $dfrac{x+60}{x+300}=dfrac{2}{5}$
