Exercise 1

Step 1

1 of 6

Remember the following rule:

$(a^m)^n=a^{mn}$

$a^n=dfrac{1}{a^n}$

We shall evaluate the following expressions.

Step 2

2 of 6

a) $25^{1/2}+16^{3/4}$

$=(5^2)^{1/2}+(2^4)^{3/4}$

$=5+2^3$

$=5+8$

$$
=13
$$

Factor the base as:

$25=5^2$

$$
16=2^4
$$

Step 3

3 of 6

b) $8^{2/3}-81^{3/4}+4^2$

$=(2^3)^{2/3}-(3^4)^{3/4}+4^2$

$=2^2-3^3+4^2$

$$
=-7
$$

Factor the base as:

$8=2^3$

$$
81=3^4
$$

Step 4

4 of 6

c) $8^{-3/4}-81^{-3/4}+8^{-3}$

$=(2^3)^{-3/4}-(3^4)^{-3/4}+8^{-3}$

$=2^{-9/4}-3^{-3}+8^{-3}$

$$
approx 0.175
$$

Factor the base as:

$8=2^3$

$$
81=3^4
$$

Step 5

5 of 6

d) $81^{-3/4}+16^{-3/4}-16^{-1/2}$

$=(3^4)^{-3/4}+(2^4)^{-3/4}-(4^2)^{-1/2}$

$=3^{-3}+2^{-3}-4^{-1}$

$$
approx -0.0879
$$

Factor the base as:

$81=3^4$

$16=2^4=4^2$

Result

6 of 6

option (b)

Exercise 2

Step 1

1 of 5

Substitute $x=2$

a) $3^{2x-1}=3^{2(2)-1}=3^{3}=27$

The statement is TRUE.

Step 2

2 of 5

b) $6^{2x-3}=6^{2(2)-3}=6^{1}=6neq sqrt{6}$

The statement is FALSE.

Step 3

3 of 5

c) $5^{3x+2}=5^{3(2)+2}=5^{8}neq dfrac{1}{5}$

The statement is FALSE.

Step 4

4 of 5

d) $(2^{2x})(2^{x-1})=2^{2(2)}(2^{2-1})=2^4(2)=16(2)=32$

The statement is TRUE.

Result

5 of 5

options (a) and (d)

Exercise 3

Step 1

1 of 5

[begin{gathered}
a) left( {{a^{10 + 2p}}} right)left( {{a^{ – p – 8}}} right) hfill \\
= {a^{left( {10 + 2p} right) – left( { – p – 8} right)}} hfill \\
= {a^{10 + 2p + p + 8}} hfill \\
= {a^{3p + 18}} hfill \\
end{gathered} ]

Remember the given rule

$$
a^m times a^n=a^{m+n}
$$

Step 2

2 of 5

[begin{gathered}
b) {left( {2{x^2}} right)^{3 – 2m}}{left( {frac{1}{x}} right)^{2m}} hfill \\
= {left( {2{x^2}} right)^{3 – 2m}}{x^{ – 2m}} hfill \\
= {2^{3 – 2m}} cdot {x^{2left( {3 – 2m} right)}} cdot {x^{ – 2m}} hfill \\
= {2^{3 – 2m}} cdot {x^{6 – 4m}} cdot {x^{ – 2m}} hfill \\
= {2^{3 – 2m}} cdot {x^{6 – 4m – 2m}} hfill \\
= {2^{3 – 2m}} cdot {x^{6 – 6m}} hfill \\
end{gathered} ]

$(ab)^m=a^mtimes b^m$

$$
dfrac{a^m}{a^n}=a^{m-n}
$$

Step 3

3 of 5

[begin{gathered}
c) left[ {{{left( c right)}^{2n – 3m}}} right]{left( {{c^3}} right)^m} div {left( {{c^2}} right)^n} hfill \
hfill \
= frac{{left[ {{{left( c right)}^{2n – 3m}}} right]{{left( {{c^3}} right)}^m}}}{{{{left( {{c^2}} right)}^n}}} hfill \
hfill \
= frac{{{c^{2n – 3m}} cdot {c^{3m}}}}{{{c^{2n}}}} hfill \
hfill \
= {c^{2n – 3m}} cdot {c^{3m}} cdot {c^{ – 2n}} hfill \
hfill \
= {c^{2n – 2n – 3m + 3m}} hfill \
hfill \
= {c^0} hfill \
hfill \
= 1 hfill \
end{gathered} ]

$(ab)^m=a^mtimes b^m$

$$
dfrac{a^m}{a^n}=a^{m-n}
$$

Step 4

4 of 5

[begin{gathered}
d) {left[ {left( {{x^{4n – m}}} right)left( {frac{1}{x}} right)} right]^6} hfill \
hfill \
= {x^{4n – m}} cdot {x^{ – 6}} hfill \
hfill \
= {x^{4n – m – 6}} hfill \
end{gathered} ]

$(ab)^m=a^mtimes b^m$

$$
dfrac{a^m}{a^n}=a^{m-n}
$$

Result

5 of 5

option (c)

Exercise 4

Step 1

1 of 2

Since the population grows $5%$ per year, it can be modeled as exponential function

$A=P(1.05)^t$

where

$P =$ population in the year 2000

$A$ = the population after $t$ years

At $t=20$

$$
A=15,000(1.05)^{20}=40,000
$$

Result

2 of 2

option (a)

Exercise 5

Step 1

1 of 3

Step 2

2 of 3

In this case,

The terminal point is $(-7,24)$ at Quadrant II

We can solve $r$ using Pythagorean relation $c^2=a^2+b^2$

$r=sqrt{(-7)^2+24^2}=25$

Substituting the value of $r$

$sin alpha =dfrac{24}{25} implies alpha= sin^{-1}left(dfrac{24}{25}right)approx 74^circ$

The acute angle is $74^circ$

The principal angle is $theta=180-74=106^circ$

Result

3 of 3

option (a)

Exercise 6

Step 1

1 of 3

begin{table}[]
defarraystretch{2.5}%
begin{tabular}{llll}
hline
multicolumn{1}{|l|}{begin{tabular}[c]{@{}l@{}}Principal \ Angle $theta$end{tabular}} & multicolumn{1}{l|}{Quadrant} & multicolumn{1}{l|}{begin{tabular}[c]{@{}l@{}}Sign of\ Coordinatesend{tabular}} & multicolumn{1}{l|}{begin{tabular}[c]{@{}l@{}}Reference\ Acute angleend{tabular}} \ hline
multicolumn{1}{|l|}{$0^circ < theta< 90^circ$} & multicolumn{1}{l|}{I} & multicolumn{1}{l|}{$(+,+)$} & multicolumn{1}{l|}{$theta$} \ hline
multicolumn{1}{|l|}{$90^circ < theta <180^circ$} & multicolumn{1}{l|}{II} & multicolumn{1}{l|}{$(-,+)$} & multicolumn{1}{l|}{$180^circ-theta$} \ hline
multicolumn{1}{|l|}{$180^circ< theta < 270^circ$} & multicolumn{1}{l|}{III} & multicolumn{1}{l|}{$(-,-)$} & multicolumn{1}{l|}{$theta-180^circ$} \ hline
multicolumn{1}{|l|}{$270^circ< theta < 360^circ$} & multicolumn{1}{l|}{IV} & multicolumn{1}{l|}{$(+,-)$} & multicolumn{1}{l|}{$360^circ-theta$} \ hline
& & & \ cline{1-3}
multicolumn{1}{|l|}{begin{tabular}[c]{@{}l@{}}Reference\ Acute angleend{tabular}} & multicolumn{1}{l|}{cosine} & multicolumn{1}{l|}{sine} & \ cline{1-3}
multicolumn{1}{|l|}{30$^circ$} & multicolumn{1}{l|}{$dfrac{sqrt{3}}{2}$} & multicolumn{1}{l|}{$dfrac{1}{2}$} & \ cline{1-3}
multicolumn{1}{|l|}{45$^circ$} & multicolumn{1}{l|}{$dfrac{sqrt{2}}{2}$} & multicolumn{1}{l|}{$dfrac{sqrt{2}}{2}$} & \ cline{1-3}
multicolumn{1}{|l|}{60$^circ$} & multicolumn{1}{l|}{$dfrac{1}{2}$} & multicolumn{1}{l|}{$dfrac{sqrt{3}}{2}$} & \ cline{1-3}
& & &
end{tabular}
end{table}

Step 2

2 of 3

To evaluate $csc 300^circ$, we must evaluate $sin 300^circ$

Determine which quadrant does $300^circ$ lie to obtain reference angle.

From the table above, $300^circ$ lies in QIV where the reference acute angle is

$360^circ-300^circ=60^circ$

At quadrant IV, $sin theta < 0$, thus

$sin 300^circ=-sin 60^circ= -dfrac{sqrt{3}}{2}$

Using the identity $csc theta =dfrac{1}{sin theta}$

$$
csc 300^circ =dfrac{1}{-sqrt{3}/2}=-dfrac{2}{sqrt{3}}=-dfrac{2sqrt{3}}{3}
$$

Result

3 of 3

option (c)

Exercise 7

Step 1

1 of 5

[begin{gathered}
{mathbf{a)}} hfill \
hfill \
left( {1 – {{tan }^2}theta } right)left( {1 – {{cos }^2}theta } right) hfill \
= left( {1 – frac{{{{sin }^2}theta }}{{{{cos }^2}theta }}} right)left( {{{sin }^2}theta } right) hfill \
= frac{{{{sin }^2}theta – {{sin }^4}theta }}{{{{cos }^2}theta }} hfill \
= frac{{{{sin }^2}theta – {{sin }^4}theta }}{{{{cos }^2}theta }} hfill \
= frac{{{{sin }^2}theta – {{sin }^4}theta }}{{1 – {{sin }^2}theta }} hfill \
hfill \
{text{The equation is an identity}}{text{.}} hfill \
end{gathered} ]

[begin{gathered}
{text{Use the following identities}} hfill \
hfill \
{text{tan}}theta = frac{{sin theta }}{{cos theta }}{text{ }} hfill \
hfill \
{text{ si}}{{text{n}}^2}theta + {cos ^2}theta = 1 hfill \
end{gathered} ]

Step 2

2 of 5

$bold{b}$)

$tan x+sin xneq tan x sin x$

for some value of $x$,

For example, when $x=45^circ$

$tan x+sin x=1+dfrac{sqrt{2}}{2}$

$tan x sin x =dfrac{sqrt{2}}{2}$

Thus, this is NOT an identity.

Disprove by counterexample.

Step 3

3 of 5

[begin{gathered}
{mathbf{c}}) hfill \
= 1 – tan theta hfill \
= 1 – frac{{sin theta }}{{cos theta }} hfill \
= frac{{cos theta – sin theta }}{{cos theta }} cdot frac{{cos theta + sin theta }}{{cos theta + sin theta }} hfill \
= frac{{{{cos }^2}theta – {{sin }^2}theta }}{{{{cos }^2}theta + sin theta cos theta }} hfill \
end{gathered} ]

Use the identity

$$
tan theta=dfrac{sintheta}{costheta}
$$

Step 4

4 of 5

[begin{gathered}
{mathbf{d}}) hfill \
hfill \
frac{{left( {frac{{1 – cos x}}{{cos x}}} right)}}{{tan x}} hfill \
= left( {frac{{1 – cos x}}{{cos x}}} right)left( {frac{{cos x}}{{sin x}}} right) hfill \
= frac{{1 – cos x}}{{sin x}} hfill \
end{gathered} ]

Use the identity

$$
tan theta=dfrac{sintheta}{costheta}
$$

Result

5 of 5

option (b)

Exercise 8

Step 1

1 of 3

In this case,

$theta = 63^circ$

opposite side = $12$

adjacent side = $x$

$tan 63^circ=dfrac{12}{x}$

$$
x=dfrac{12}{tan 63^circ}approx 6
$$

Remember that for right triangles

$$
tan theta =dfrac{text{opposite;side}}{text{adjacent side}}
$$

Exercise scan

Step 2

2 of 3

$dfrac{sin 63^circ}{12}=dfrac{sin 27^circ}{x}$

$x=dfrac{12sin 27^circ}{sin 63^circ}approx 6$

In both methods, the answer is option (c)

Alternatively, you can use sine law.

The sum of the angles in a triangle is always $180^circ$, so the other acute angle must be $180-90-63=27^circ$

The formula is

$$
dfrac{sin alpha}{A}=dfrac{sin beta}{B}=dfrac{sin gamma}{C}
$$

Exercise scan

Result

3 of 3

option (c)

Exercise 9

Step 1

1 of 2

In this case, the opposite side
and hypotenuse is given,
so we must use sine function.

opposite side = $7$

hypotenuse = $22$

$sin theta = dfrac{7}{22}$

$theta=sin^{-1}left(dfrac{7}{22}right)$

$theta approx 19^circ$

The answer is option (a)

Remember that for right triangles

$sin theta =dfrac{text{opposite;side}}{text{hypotenuse}}$

$cos theta= dfrac{text{adjacent;side}}{text{hypotenuse}}$

$$
tan theta= dfrac{text{opposite side}}{text{adjacent side}}
$$

Exercise scan

Result

2 of 2

option (a)

Exercise 10

Step 1

1 of 2

In the figure, we are given with the adjacent side and hypotenuse.

Since we want to know $csc theta = dfrac{1}{sin theta}$,

we need to solve for the opposite side, which can be done using Pythagorean relation.

$c^2=a^2+b^2$

opposite side = $sqrt{13^2-5^2}=12$

Therefore,

$csc theta = dfrac{1}{sin theta}=dfrac{1}{(12/13)}=dfrac{13}{12}$

The answer is option (c)

Remember that for right triangles

$sin theta =dfrac{text{opposite;side}}{text{hypotenuse}}$

$cos theta= dfrac{text{adjacent;side}}{text{hypotenuse}}$

$$
tan theta= dfrac{text{opposite side}}{text{adjacent side}}
$$

Exercise scan

Result

2 of 2

option (c)

Exercise 11

Step 1

1 of 2

You can use this identity

$sec^2theta=tan^2theta+1$

$sectheta=sqrt{tan^2theta+1}$

$sec theta=sqrt{left(dfrac{4}{3}right)^2+1}$

$sec theta=dfrac{5}{3}$

Since $costheta=dfrac{1}{sectheta}=dfrac{1}{5/3}=dfrac{3}{5}$

In quadrant III, $costheta<0$, therefore

$cos theta=-dfrac{3}{5}$

The answer is option (b)

Result

2 of 2

option (b)

Exercise 12

Step 1

1 of 4

[begin{gathered}
{text{Remember that}} hfill \
sin theta = frac{{{text{opposite side}}}}{{{text{hypotenuse}}}} hfill \
cos theta = frac{{{text{adjacent side}}}}{{{text{hypotenuse}}}} hfill \
{text{tan}}theta = frac{{{text{opposite side}}}}{{{text{adjacent side}}}} hfill \
end{gathered} ] Exercise scan

Step 2

2 of 4

Step 3

3 of 4

[begin{gathered}
tan {57^ circ } = frac{y}{{15 + x}} hfill \
tan {83^ circ } = frac{y}{x} Rightarrow x = frac{y}{{tan {{83}^ circ }}} hfill \
hfill \
{text{Substitute }}x{text{ to the first equation}} hfill \
hfill \
tan {57^ circ } = frac{y}{{15 + x}} hfill \
tan {57^ circ } = frac{y}{{15 + left( {frac{y}{{tan {{83}^ circ }}}} right)}} hfill \
1.54 = frac{y}{{15 + 0.123y}} hfill \
1.54left( {15 + 0.123y} right) = y hfill \
23.1 + 0.1894y = y hfill \
hfill \
y = frac{{23.1}}{{1 – 0.1894}} = boxed{28.5{text{ km}}} hfill \
hfill \
{text{option (a)}} hfill \
end{gathered} ]

Result

4 of 4

option (a)

Exercise 13

Step 1

1 of 4

[begin{gathered}
{text{Remember that}} hfill \
sin theta = frac{{{text{opposite side}}}}{{{text{hypotenuse}}}} hfill \
cos theta = frac{{{text{adjacent side}}}}{{{text{hypotenuse}}}} hfill \
{text{tan}}theta = frac{{{text{opposite side}}}}{{{text{adjacent side}}}} hfill \
end{gathered} ] Exercise scan

Step 2

2 of 4

Step 3

3 of 4

[begin{gathered}
{text{In this case,}} hfill \
theta = {35^ circ } hfill \
{text{opposite side = 12m}} hfill \
{text{adjacent side = }}x hfill \
hfill \
sin theta = frac{{12}}{x} hfill \
x = frac{{12}}{{sin {{35}^ circ }}} = boxed{20.9{text{ m}}} hfill \
hfill \
{text{The answer is option (a)}} hfill \
end{gathered} ]

Result

4 of 4

option (a)

Exercise 14

Step 1

1 of 4

$$
begin{gathered}
{text{Remember the formula for sine law}} \
\
frac{{sin alpha }}{a} = frac{{sin beta }}{b} = frac{{sin gamma }}{c} \
end{gathered}
$$

Exercise scan

Step 2

2 of 4

Step 3

3 of 4

[begin{gathered}
frac{{sin {{32}^{text{o}}}}}{{24.1}} = frac{{sin {text{8}}{{text{1}}^{text{o}}}}}{{AB}} Rightarrow AB = frac{{24.1sin {{81}^{text{o}}}}}{{sin {{32}^{text{o}}}}} approx 44.9 hfill \
hfill \
angle B = 180 – left( {32 + 81} right) = {67^{text{o}}} hfill \
hfill \
frac{{sin {{67}^{text{o}}}}}{{AC}} = frac{{sin {{32}^{text{o}}}}}{{24.1}} Rightarrow AC = frac{{24.1sin {{67}^{text{o}}}}}{{sin {{32}^{text{o}}}}} = 41.9 hfill \
hfill \
{text{Therefore, the answer is option }}left( {text{d}} right) hfill \
end{gathered} ]

Result

4 of 4

option (d)

Exercise 15

Step 1

1 of 3

[begin{gathered}
{text{For a sinusoidal function, remember the following }} hfill \
hfill \
y = Acos left[ {kleft( {x – d} right)} right] + c{text{ }} hfill \
{text{or }} hfill \
y = Asin left[ {kleft( {x – d} right)} right] + c{text{ }} hfill \
hfill \
{text{amplitude = }}|A| = dfrac{|y_{text{max}}-y_{text{min}}|}{2}hfill \
{text{period = }}T=frac{{{{360}^ circ }}}{|k|} hfill \
{text{equation of axis:}}{text{ }},,,y = c = dfrac{y_{text{max}}+y_{text{min}}}{2} hfill \
hfill \
{text{Generally, if the maximum point is translated $d$ units to the right of the}}hfill \\
{text{$y$-axis, the equation is $y=Acos[k(x-d)^circ]+c$}}hfill \\
{text{If the minimum point is translated $d$ units to the right of the}}hfill \\
{text{$y$-axis, the equation is $y=-Acos[k(x-d)^circ]+c$}}hfill \\
{text{Note that multiple possible sinusoidal functions can describe a graph or data}}hfill \
end{gathered} ]

Step 2

2 of 3

From the given function $y=2cos2(theta+45^circ)$, notice that the maximum point must be shifted $45^circ$ to the left of $y$-axis.

From the given options, the only one that satisfies this is option (d)

Result

3 of 3

option (d)

Exercise 16

Step 1

1 of 3

[begin{gathered}
{text{For a sinusoidal function, remember the following }} hfill \
hfill \
y = Acos left[ {kleft( {x – d} right)} right] + c{text{ }} hfill \
{text{or }} hfill \
y = Asin left[ {kleft( {x – d} right)} right] + c{text{ }} hfill \
hfill \
{text{amplitude = }}|A| = dfrac{|y_{text{max}}-y_{text{min}}|}{2}hfill \
{text{period = }}T=frac{{{{360}^ circ }}}{|k|} hfill \
{text{equation of axis:}}{text{ }},,,y = c = dfrac{y_{text{max}}+y_{text{min}}}{2} hfill \
hfill \
{text{Generally, if the maximum point is translated $d$ units to the right of the}}hfill \\
{text{$y$-axis, the equation is $y=Acos[k(x-d)^circ]+c$}}hfill \\
{text{If the minimum point is translated $d$ units to the right of the}}hfill \\
{text{$y$-axis, the equation is $y=-Acos[k(x-d)^circ]+c$}}hfill \\
{text{Note that multiple possible sinusoidal functions can describe a graph or data}}hfill \
end{gathered} ]

Step 2

2 of 3

In the given function ,$y=2cos(2theta)$, we expect that the maximum point must be at the $y$-axis. Among the given options, only option (a) satisfies this.

Result

3 of 3

option (a)

Exercise 17

Step 1

1 of 3

[begin{gathered}
{text{For a sinusoidal function, remember the following }} hfill \
hfill \
y = Acos left[ {kleft( {x – d} right)} right] + c{text{ }} hfill \
{text{or }} hfill \
y = Asin left[ {kleft( {x – d} right)} right] + c{text{ }} hfill \
hfill \
{text{amplitude = }}|A| = dfrac{|y_{text{max}}-y_{text{min}}|}{2}hfill \
{text{period = }}T=frac{{{{360}^ circ }}}{|k|} hfill \
{text{equation of axis:}}{text{ }},,,y = c = dfrac{y_{text{max}}+y_{text{min}}}{2} hfill \
hfill \
{text{Generally, if the maximum point is translated $d$ units to the right of the}}hfill \\
{text{$y$-axis, the equation is $y=Acos[k(x-d)^circ]+c$}}hfill \\
{text{If the minimum point is translated $d$ units to the right of the}}hfill \\
{text{$y$-axis, the equation is $y=-Acos[k(x-d)^circ]+c$}}hfill \\
{text{Note that multiple possible sinusoidal functions can describe a graph or data}}hfill \
end{gathered} ]

Step 2

2 of 3

In this case, $T=720^circ$ so $k=dfrac{360}{720}=0.5$

The range is given so we can obtain the axis halfway between the minimum and maximum,

$y=dfrac{2+12}{2}=7$

From the given options, only option (c) satisfies this.

Result

3 of 3

option (c)

Exercise 18

Step 1

1 of 5

a) The amplitude affects the maximum and minimum values as well as the range. The statement is TRUE.

Step 2

2 of 5

b) Changing value of $k$ does NOT affect amplitude and axis, but only affects the period. The statement is FALSE.

Step 3

3 of 5

c) Changing the value of $c$ only affects the axis and does NOT affect the period and amplitude. The statement is FALSE.

Step 4

4 of 5

d) Changing value of $d$ does NOT affect the period, amplitude, and axis. The statement is FALSE.

Result

5 of 5

option (a)

Exercise 19

Step 1

1 of 4

Step 2

2 of 4

You can visualize the problem as shown above.

The woman’s coordinate after $18$ minutes, is

$(x,y)implies (20cos(7.5times 18)^circ,20sin(7.5times 18)^circ)=(-10sqrt{2}.10sqrt{2})$

$sin alpha = dfrac{10sqrt{2}}{20}implies alpha =sin^{-1}left( dfrac{10sqrt{2}}{20}right)=45^circ$

$$
theta=180-45=135^circ
$$

Step 3

3 of 4

Now, you can use cosine law to find the shortest distance $x$

$c^2=a^2+b^2-2abcostheta$

$x^2=20^2+20^2-2(20)(20)cos 135^circ$

$x=sqrt{20^2+20^2-2(20)(20)cos 135^circ}approx 37$

The answer is option (b)

Result

4 of 4

option (b)

Exercise 20

Step 1

1 of 2

For $y=sin x$

a) The period of any sine function $y=Asin[k(x-d)]+c$ is $T=dfrac{360}{|k|}$. Since $k=1$ in this case, its period is $360^circ$ so the statement is TRUE.

b) The amplitude of any sine function $y=Asin[k(x-d)]+c$ is $T=dfrac{360}{|k|}$. Since $A=1$, the amplitude is 1 and the statement is true.

c) The equation of axis of any sine function $y=Asin[k(x-d)]+c$ is $y=c$. Since $c=0$, the statement is true.

d) The range of any sine function $y=Asin[k(x-d)]+c$ is
${ y in bold{R};|;c-|A|leq y leq c+|A|}$.
Thus, the range of $y=sin x$ is ${ y in bold{R};|;-1leq y leq 1}$, so the statement is FALSE.

Result

2 of 2

option (d)

Exercise 21

Step 1

1 of 3

Step 2

2 of 3

You can visualize the problem by sketching. The octagon is inscribed in a circle.

To find the perimeter of the regular octagon, we must find the length of each side and multiply by 8.

The angle is $theta =dfrac{360}{8}=45^circ$

The length of each side is denoted as $x$ in the figure which can be solved by cosine law.

$x=sqrt{14^2+14^2-2(14)(14)cos 45^circ}approx 10.715$

Perimeter $=8times 10.715=85.72$ cm

The answer is option (c)

Result

3 of 3

option (c)

Exercise 22

Step 1

1 of 5

Sketch the triangle as follows to help you visualize the problem Exercise scan

Step 2

2 of 5

[begin{gathered}
{text{The sum of the angles in a triangle must always be 18}}{{text{0}}^{text{o}}} hfill \
{text{Since one angle is 8}}{{text{5}}^{text{o}}},{text{ the sum of the other angles must be 9}}{{text{5}}^{text{o}}} hfill \
{text{Let }}angle {text{C}} = theta {text{, }}angle {text{C}} = {95^{text{o}}} – theta hfill \
{text{Remember that sin}}theta = frac{{{text{opposite side}}}}{{{text{hypotenuse}}}} hfill \
{text{sin}}theta {text{ = }}frac{x}{{15}}{text{ ; sin}}left( {{{95}^{text{o}}} – theta } right){text{ = }}frac{x}{{10}} hfill \
{text{ }} hfill \
{text{Equate }}x hfill \
x = 15sin theta = 10sin left( {{{95}^{text{o}}} – theta } right) hfill \
{text{We shall solve this graphically}} hfill \
end{gathered} ]

Step 3

3 of 5

Step 4

4 of 5

Since $theta=35.188$, the height of the triangle is

$x=15sin(35.188)approx 8.64$

The answer is option (b)

Result

5 of 5

option (b)

Exercise 23

Step 1

1 of 4

To evaluate the value of a trigonometric function, you can perform the following steps:

(1) Determine which quadrant does the terminal point lie using the table below. If the given angle $beta$ is outside $0leq thetaleq 360^circ$, find a coterminal angle $theta$ within this range using the formula

coterminal angle = $beta+360^circ n$ where $n$ is an integer.

(2) Calculate the reference acute angle based on its quadrant.

(3) Find the absolute value of sine or cosine based on the reference angle as shown in the table.

(4) Change the sign of the sine or cosine based on its quadrant.

(5) Use the identities to find the value of other trigonometric functions.

$tan = dfrac{sin theta}{cos theta}$

$sec theta = dfrac{1}{cos theta}$

$csc theta = dfrac{1}{sin theta}$

$cot theta=dfrac{1}{cottheta}$

Step 2

2 of 4

begin{table}[]
center
defarraystretch{2.4}%
begin{tabular}{llll}
hline
multicolumn{1}{|l|}{begin{tabular}[c]{@{}l@{}}principal angle;$theta$end{tabular}} & multicolumn{1}{l|}{Quadrant} & multicolumn{1}{l|}{begin{tabular}[c]{@{}l@{}}sign $(cos theta, sin theta)$end{tabular}} & multicolumn{1}{l|}{begin{tabular}[c]{@{}l@{}}reference acute angle $alpha$end{tabular}} \ hline
multicolumn{1}{|l|}{$0< theta<90^circ$} & multicolumn{1}{l|}{I} & multicolumn{1}{l|}{$(+,+)$} & multicolumn{1}{l|}{$theta$} \ hline
multicolumn{1}{|l|}{$90^circ <theta < 180^circ$} & multicolumn{1}{l|}{II} & multicolumn{1}{l|}{$(-,+)$} & multicolumn{1}{l|}{$180^circ-theta$} \ hline
multicolumn{1}{|l|}{$180^circ <theta < 270^circ$} & multicolumn{1}{l|}{III} & multicolumn{1}{l|}{$(-,-)$} & multicolumn{1}{l|}{$theta-180^circ$} \ hline
multicolumn{1}{|l|}{$270^circ < theta <360^circ$} & multicolumn{1}{l|}{IV} & multicolumn{1}{l|}{$(+,-)$} & multicolumn{1}{l|}{$360^circ-theta$} \ hline
& & & \ cline{1-3}
multicolumn{1}{|l|}{begin{tabular}[c]{@{}l@{}}reference acute angle $alpha$ end{tabular}} & multicolumn{1}{l|}{$cos alpha$} & multicolumn{1}{l|}{$sin alpha$} & \ cline{1-3}
multicolumn{1}{|l|}{30$^circ$} & multicolumn{1}{l|}{$dfrac{sqrt{3}}{2}$} & multicolumn{1}{l|}{$dfrac{1}{2}$} & \ cline{1-3}
multicolumn{1}{|l|}{45$^circ$} & multicolumn{1}{l|}{$dfrac{sqrt{2}}{2}$} & multicolumn{1}{l|}{$dfrac{sqrt{2}}{2}$} & \ cline{1-3}
multicolumn{1}{|l|}{60$^circ$} & multicolumn{1}{l|}{$dfrac{1}{2}$} & multicolumn{1}{l|}{$dfrac{sqrt{3}}{2}$} & \ cline{1-3}
& & &
end{tabular}
end{table}

Step 3

3 of 4

$-420^circ$ is outside the range of $0leq theta leq 360^circ$, so we must find its coterminal angle within the range.

coterminal angle = $-420+360(2)=300^circ$

$300^circ$ lies within $270^circ < theta 0$ in QIV, $cos(-420^circ)=dfrac{1}{2}$

Result

4 of 4

$dfrac{1}{2}$

Exercise 24

Step 1

1 of 3

[begin{gathered}
frac{{{text{si}}{{text{n}}^2}theta + {{cos }^2}theta }}{{left( {frac{{cos theta }}{{sin theta }}} right)}} = left( {{{sin }^2}theta + {{cos }^2}theta } right)left( {frac{{sin theta }}{{cos theta }}} right) hfill \
hfill \
{text{Use the identity }}{sin ^2}theta + {cos ^2}theta = 1,,,,,,{text{and }},,{text{tan}}theta = frac{{sin theta }}{{cos theta }} hfill \
hfill \
= left( {{{sin }^2}theta + {{cos }^2}theta } right)left( {frac{{sin theta }}{{cos theta }}} right) hfill \
= 1 cdot tan theta hfill \
= frac{y}{x} hfill \
end{gathered} ]

Step 2

2 of 3

The answer is option $(b)$

Result

3 of 3

option (b)

Exercise 25

Step 1

1 of 3

[begin{gathered}
frac{{sin x cdot sin x}}{{left( {1 – sin x} right)left( {1 + sin x} right)}} hfill \
hfill \
{text{Remember that }}left( {a – b} right)left( {a + b} right) = {a^2} – {b^2} hfill \
= frac{{{{sin }^2}x}}{{1 – {{sin }^2}x}} hfill \
hfill \
{text{Use the identity }}{cos ^2}x = 1 – {sin ^2}x hfill \
hfill \
= frac{{{{sin }^2}x}}{{{{cos }^2}x}} = {left( {frac{{sin x}}{{cos x}}} right)^2} hfill \
hfill \
{text{Use }}tan x = frac{{sin x}}{{cos x}} hfill \
hfill \
= {tan ^2}x hfill \
end{gathered} ]

Step 2

2 of 3

The answer is option $(c)$

Result

3 of 3

option (c)

Exercise 26

Step 1

1 of 3

[begin{gathered}
{text{For a sinusoidal function, remember the following }} hfill \
hfill \
y = Acos left[ {kleft( {x – d} right)} right] + c{text{ }} hfill \
{text{or }} hfill \
y = Asin left[ {kleft( {x – d} right)} right] + c{text{ }} hfill \
hfill \
{text{amplitude = }}|A| = dfrac{|y_{text{max}}-y_{text{min}}|}{2}hfill \
{text{period = }}T=frac{{{{360}^ circ }}}{|k|} hfill \
{text{equation of axis:}}{text{ }},,,y = c = dfrac{y_{text{max}}+y_{text{min}}}{2} hfill \
hfill \
end{gathered} ]

Step 2

2 of 3

For $y=sin 4theta$

period $=T=dfrac{360^circ}{|4|}=90^circ$

The answer is option (c)

Result

3 of 3

option (c)

Exercise 27

Step 1

1 of 3

Remember that $dfrac{1}{x^{-m}}=x^m$ and $left(dfrac{x}{y}right)^{-1}=dfrac{y}{x}$

Step 2

2 of 3

[begin{gathered}
{left[ {left( {frac{1}{a}} right)left( {frac{1}{{{b^{ – 1}}}}} right)} right]^{ – 1}} hfill \
= {left[ {frac{1}{a}left( b right)} right]^{ – 1}} hfill \
= {left( {frac{b}{a}} right)^{ – 1}} hfill \
= left( {frac{a}{b}} right) hfill \
end{gathered} ]

Result

3 of 3

$$
dfrac{a}{b}
$$

Exercise 28

Step 1

1 of 3

$3x^{1/2}=12$

$dfrac{3x^{1/2}}{3}=dfrac{12}{3}$

$$
x^{1/2}=4
$$

Divide both sides by $3$

Step 2

2 of 3

$(x^{1/2})^2=4^2$

$x=16$

The answer is option (c)

Square both sides

Result

3 of 3

option (c)

Exercise 29

Step 1

1 of 4

The thickness of a letter paper is $0.1$ mm thick. In each fold, the thickness doubles. Thus, we can model the thickness of $n$th fold,

$t_n=0.1times 2^n$

553 m $times dfrac{1000;text{mm}}{1;text{m}}=0.1times 2^n$

We can find the value of $m$ graphically.

Step 2

2 of 4

Step 3

3 of 4

Therefore, it would reach 55$3$ m on the 23rd fold.

Result

4 of 4

23rd fold

Exercise 30

Step 1

1 of 3

We are given two included sides $AC=75$ and $AB=55$ and a set of included angle $angle A=105^circ, 125^circ,145^circ, 165^circ, 175^circ$

For each angle, we shall find $BC$

Step 2

2 of 3

The cosine law for this case is

$BC^2=AC^2+AB^2-2(AC)(AB)cos angle A$

$angle A=105^circ$ : $BC=sqrt{75^2+55^2-2(75)(55)cos(105^circ)}=103.85$ cm

$angle A=125^circ$ : $BC=sqrt{75^2+55^2-2(75)(55)cos(125^circ)}=115.68$ cm

$angle A=145^circ$ : $BC=sqrt{75^2+55^2-2(75)(55)cos(145^circ)}=124.13$ cm

$angle A=165^circ$ : $BC=sqrt{75^2+55^2-2(75)(55)cos(165^circ)}=128.91$ cm

$angle A=175^circ$ : $BC=sqrt{75^2+55^2-2(75)(55)cos(175^circ)}=129.88$ cm

Result

3 of 3

See answers inside.

Exercise 31

Step 1

1 of 5

a) The graph can be obtained from the given data as follows. Exercise scan

Step 2

2 of 5

[begin{gathered}
{text{For a sinusoidal function, remember the following }} hfill \
hfill \
y = Acos left[ {kleft( {x – d} right)} right] + c{text{ }} hfill \
{text{or }} hfill \
y = Asin left[ {kleft( {x – d} right)} right] + c{text{ }} hfill \
hfill \
{text{amplitude = }}|A| = dfrac{|y_{text{max}}-y_{text{min}}|}{2}hfill \
{text{period = }}T=frac{{{{360}^ circ }}}{|k|} hfill \
{text{equation of axis:}}{text{ }},,,y = c = dfrac{y_{text{max}}+y_{text{min}}}{2} hfill \
hfill \
{text{Generally, if the maximum point is translated $d$ units to the right of the}}hfill \\
{text{$y$-axis, the equation is $y=Acos[k(x-d)^circ]+c$}}hfill \\
{text{If the minimum point is translated $d$ units to the right of the}}hfill \\
{text{$y$-axis, the equation is $y=-Acos[k(x-d)^circ]+c$}}hfill \\
{text{Note that multiple possible sinusoidal functions can describe a graph or data}}hfill \
end{gathered} ]

Step 3

3 of 5

In this case,

$D_{text{min}}=1.25$ and $D_{text{max}}=6.25$

amplitude = $dfrac{|6.25-1.25|}{2}=2.5$

equation of axis: $D=dfrac{6.25+1.25}{2}=3.75$

period: $16-4=12implies k =dfrac{360}{12}=30$

The maximum point is shifted $4$ units to the right of $y$-axis, $implies d=4$

The equation describing the data is

$$
y=2.5cos[30(x-4)^circ]+3.75
$$

Step 4

4 of 5

b) From part(a), the maximum depth is $6.25$ m

c) Yes. The minimum depth is $1.25$ which is greater than $1$ m.

Result

5 of 5

a) $y=2.5cos[30(x-4)^circ]+3.75$

b) 6.25 m

c) Yes

All Solutions

Page 408: Cumulative Review

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