Exercise 1

Step 1

1 of 7

begin{table}[]
begin{tabular}{|l|l|}
hline
Transformation & Description \ hline
$y=f(x)+c$ & begin{tabular}[c]{@{}l@{}}vertical translation of\ $c$ units upwardend{tabular} \ hline
$y=f(x+d)$ & begin{tabular}[c]{@{}l@{}}horizontal translation of $d$ units\ to the leftend{tabular} \ hline
$y=acdot f(x)$ & begin{tabular}[c]{@{}l@{}}vertical stretching, multiplies\ the amplitude by $a$end{tabular} \ hline
$y=f(kx)$ & begin{tabular}[c]{@{}l@{}}horizontal compression, divides\ the period by $k$end{tabular} \ hline
$y=-f(x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in\ the $x$-axisend{tabular} \ hline
$y=f(-x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in \ the $y$-axisend{tabular} \ hline
end{tabular}
end{table}

Step 2

2 of 7

a) $f(x)=sin (4x)+2$

(1) horizontally compress by a factor of 4

(2) translate 2 units up

Step 3

3 of 7

b) $y=0.25cos(x-20^circ)$

(1) divide the amplitude by 4

(2) translate 20$^circ$ to the right

Step 4

4 of 7

c) $g(x)=-sin(0.5x)$

(1) reflect over $x$-axis

(2) horizontally stretch by a factor of 2

Step 5

5 of 7

d) $y=12cos(18x)+3$

(1) horizontally compress by a factor of 18

(2) vertically stretch by a factor of 12

(3) translate 3 units upward

Step 6

6 of 7

e) $f(x)=-20sinleft[dfrac{1}{3}(x-40^circ)right]$

(1) reflect over $x$-axis

(2) horizontally stretch by a factor of 3

(3) vertically stretch by a factor of 20

(4) horizontally translate 40$^circ$ to the right

Result

7 of 7

See answers inside.

Exercise 2

Step 1

1 of 3

[begin{gathered}
\
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
hfill \
{text{For a sinusoidal function}} hfill \
hfill \
y = Acos left( {kx} right) + B{text{ or }}y = Asin left( {kx} right) + B hfill \
hfill \
{text{amplitude = }}A hfill \
{text{period = }}frac{{{{360}^ circ }}}{k} hfill \
{text{equation of axis:}}{text{ }},,,y = B hfill \
hfill \
{text{If the function starts at }}x = {x_0}{text{ and goes through}},n{text{ cycles,}} hfill \
hfill \
{text{domain: }}left{ {x in {mathbf{R};|}{text{ }}{x_0} leq x leq {x_0} + nT} right} hfill \
hfill \
{text{range: }}left{ {y in {mathbf{R};|}{text{ }}{y_{{text{min}}}} leq y leq {y_{{text{max}}}}} right} hfill \
hfill \
hfill \
end{gathered} ]

Step 2

2 of 3

For $f(x)=4cos 3x+6$ starting at $x=0$ with 3 cycles

period = $dfrac{360^circ}{3}=120^circ$

amplitude = $4$

equation of the axis: $f(x)=6$

domain: ${ xin bold{R};|;0leq x leq 360^circ}$

$f(x)_{min}=6-4=2$, $f(x)_{max}=6+4=10$

range: ${ f(x) in bold{R};|;2leq f(x)leq 10}$

Result

3 of 3

period = $120^circ$

amplitude = $4$

equation of axis: $f(x)=6$

domain: ${ xin bold{R};|;0leq x leq 360^circ}$

range: ${ f(x) in bold{R};|;2leq f(x)leq 10}$

Exercise 3

Step 1

1 of 4

begin{table}[]
begin{tabular}{|l|l|}
hline
Transformation & Description \ hline
$y=f(x)+c$ & begin{tabular}[c]{@{}l@{}}vertical translation of\ $c$ units upwardend{tabular} \ hline
$y=f(x+d)$ & begin{tabular}[c]{@{}l@{}}horizontal translation of $d$ units\ to the leftend{tabular} \ hline
$y=acdot f(x)$ & begin{tabular}[c]{@{}l@{}}vertical stretching, multiplies\ the amplitude by $a$end{tabular} \ hline
$y=f(kx)$ & begin{tabular}[c]{@{}l@{}}horizontal compression, divides\ the period by $k$end{tabular} \ hline
$y=-f(x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in\ the $x$-axisend{tabular} \ hline
$y=f(-x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in \ the $y$-axisend{tabular} \ hline
end{tabular}
end{table}

Step 2

2 of 4

$g(x)=5sin(2(x-30^circ))+4$

Starting from the graph of $y=sin x$

(1) Horizontally compress by a factor of 2 (divide the period by 2)

(2) Vertically stretch by a factor of 5 (multiply the amplitude by 5)

(3) Horizontally translate $30^circ$ to the right

(4) Vertically translate 4 units upward

Step 3

3 of 4

Result

4 of 4

See graph inside.

Exercise 4

Step 1

1 of 8

begin{table}[]
begin{tabular}{|l|l|}
hline
Transformation & Description \ hline
$y=f(x)+c$ & begin{tabular}[c]{@{}l@{}}vertical translation of\ $c$ units upwardend{tabular} \ hline
$y=f(x+d)$ & begin{tabular}[c]{@{}l@{}}horizontal translation of $d$ units\ to the leftend{tabular} \ hline
$y=acdot f(x)$ & begin{tabular}[c]{@{}l@{}}vertical stretching, multiplies\ the amplitude by $a$end{tabular} \ hline
$y=f(kx)$ & begin{tabular}[c]{@{}l@{}}horizontal compression, divides\ the period by $k$end{tabular} \ hline
$y=-f(x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in\ the $x$-axisend{tabular} \ hline
$y=f(-x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in \ the $y$-axisend{tabular} \ hline
end{tabular}
end{table}

Step 2

2 of 8

a) $y=-2sin(x+10^circ)$

(1) reflect over $x$-axis

(2) vertically stretch by a factor of 2 (multiply amplitude by 2)

(3) horizontally translate $10^circ$ to the left.

Step 3

3 of 8

b) $y=cos(5x)+7$

(1) horizontally compress by a factor of 5 (divide the period by 5)

(2) vertically translate upwards by 7

Step 4

4 of 8

c) $y=9cos[2(x+6^circ)]-5$

(1) horizontally compress by a factor of 2 (divide the period by 2)

(2) vertically stretch by a factor of 9 (multiply the amplitude by 9)

(3) horizontally translate 6 units to the left

(4) vertically translate 5 units downward

Step 5

5 of 8

d) $g(x)=dfrac{1}{5}sin(x-15^circ)+1$

(1) vertically compress by a factor of 5 (divide the amplitude by 5)

(2) horizontally translate $15^circ$ to the right

(3) vertically translate 1 unit upwards

Step 6

6 of 8

e) $h(x)=-sinleft[ dfrac{1}{4}(x+37^circ)right]-2$

(1) reflect over x-axis

(2) horizontally stretch by a factor of 4 (multiply the period by 4)

(3) horizontally translate $37^circ$ to the left

(4) vertically translate 2 units downward

Step 7

7 of 8

f) $d=-6cos(3t)+22$

(1) reflect over $x$-axis

(2) horizontally compress by a factor of 3 (divide the period by 3)

(3) vertically translate 22 units upward

Result

8 of 8

See answers inside.

Exercise 5

Step 1

1 of 5

[begin{gathered}
{text{For a sinusoidal function}} 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 hfill \
{text{period = }}frac{{{{360}^ circ }}}{k} hfill \
{text{equation of axis:}}{text{ }},,,y = c hfill \
end{gathered} ]

Step 2

2 of 5

a) $y=sin(2theta -90^circ)implies y=sin[2(theta-45^circ)]$

The period must be $T=dfrac{360^circ}{2}=180^circ$

With this information, the answer is graph (ii)

Step 3

3 of 5

b) $y=sin(3theta-90^circ)implies y=sin[3(theta-30^circ)]$

The period must be $T=dfrac{360^circ}{3}=120^circ$

With this information, the answer is graph (iii)

Step 4

4 of 5

c) $y=sinleft(dfrac{theta}{2}+30^circright)implies y=sinleft[ dfrac{1}{2}(theta+15^circ)right]$

The period must be $T=dfrac{360^circ}{1/2}=720^circ$

With this information, the answer is graph (i)

Result

5 of 5

a) ii

b) iii

c) i

Exercise 6

Step 1

1 of 8

[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| hfill \
{text{period = }}T=frac{{{{360}^ circ }}}{|k|} hfill \
{text{equation of axis:}}{text{ }},,,y = c hfill \
hfill \
{text{If the function starts at }}x = {x_0}{text{ and goes through}},n{text{ cycles,}} hfill \
hfill \
{text{domain: }}left{ {x in {mathbf{R}};|;{text{ }}{x_0} leq x leq {x_0} + nT} right} hfill \
hfill \
{text{range: }}left{ {y in {mathbf{R}};|;{text{ }}c – |A| leq y leq c + |A|} right} hfill \
hfill \
hfill \
end{gathered} ]

Step 2

2 of 8

For each of the following, there are 3 cycles so $n=3$.

a) $y=3sin x +2$

period : $T=dfrac{360^circ}{1}=360^circ$

amplitude: $A=3$

equation of the axis: $y=2$

$nT=3times 360^circ=1080^circ$

$c-|A|=2-3=-1$
$c+|A|=2+3=5$

domain: ${ x in bold{R};|;0^circ leq x leq 1080^circ}$

range: ${ y in bold{R};|;-1leq y leq 5}$

Step 3

3 of 8

b) $g(x)=-4cos (2x) +7$

period : $T=dfrac{360^circ}{2}=180^circ$

amplitude: $|A|=4$

equation of the axis: $g(x)=7$

$nT=3times 180^circ=540^circ$

$c-|A|=7-4=3$
$c+|A|=7+4=11$

domain: ${ x in bold{R};|;0^circ leq x leq 540^circ}$

range: ${ g(x) in bold{R};|;3leq g(x) leq 11}$

Step 4

4 of 8

c) $h=-dfrac{1}{2}sin (t)-5$

period : $T=dfrac{360^circ}{1}=360^circ$

amplitude: $|A|=dfrac{1}{2}$

equation of the axis: $h=-5$

$nT=3times 360^circ=1080^circ$

$c-|A|=-5-dfrac{1}{2}=-dfrac{11}{2}$
$c+|A|=-5+dfrac{1}{2}=-dfrac{9}{2}$

domain: ${ t in bold{R};|;0^circ leq t leq 1080^circ}$

range: ${ h in bold{R};|;-dfrac{11}{2}leq h leq -dfrac{9}{2}}$

Step 5

5 of 8

d) $h(x)=cos(4(x-12^circ))-9$

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

amplitude: $|A|=1$

equation of the axis: $y=-9$

$nT=3times 90^circ=270^circ$

$c-|A|=-9-1=-10$
$c+|A|=-9+1=-8$

domain: ${ t in bold{R};|;0^circ leq x leq 270^circ}$

range: ${ h(x) in bold{R};|;-10leq h(x) leq -8}$

Step 6

6 of 8

e) $d=10sinleft(180(t-17^circ)right)-30$

period : $T=dfrac{360^circ}{180}=2^circ$

amplitude: $|A|=10$

equation of the axis: $d=-30$

$nT=3times 2^circ=6^circ$

$c-|A|=-30-10=-40$
$c+|A|=-30+10=-20$

domain: ${ t in bold{R};|;0^circleq x leq 6^circ}$

range: ${ d in bold{R};|;-40leq d leq -20}$

Step 7

7 of 8

f) $j(x)=0.5sinleft(2x-30^circright)$

period : $T=dfrac{360^circ}{2}=180^circ$

amplitude: $|A|=0.5$

equation of the axis: $j(x)=0$

$nT=3times 180^circ=540^circ$

$c-|A|=0-0.5=-0.5$
$c+|A|=0+0.5=0.5$

domain: ${ t in bold{R};|;0^circleq x leq 540^circ}$

range: ${ j(x) in bold{R};|;-0.5leq j(x) leq 0.5}$

Result

8 of 8

See explanation inside.

Exercise 7

Step 1

1 of 8

[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| hfill \
{text{period = }}T=frac{{{{360}^ circ }}}{|k|} hfill \
{text{equation of axis:}}{text{ }},,,y = c hfill \
hfill \
{text{If the function starts at }}x = {x_0}{text{ and goes through}},n{text{ cycles,}} hfill \
hfill \
{text{domain: }}left{ {x in {mathbf{R}};|;{text{ }}{x_0} leq x leq {x_0} + nT} right} hfill \
hfill \
{text{range: }}left{ {y in {mathbf{R}};|;{text{ }}c – |A| leq y leq c + |A|} right} hfill \
hfill \
hfill \
end{gathered} ]

Step 2

2 of 8

a) $y=2sin x$+3

(1) vertically stretch by a factor of 2 (multiply amplitude by 2)

(2) Translate $2$ units upward

Exercise scan

Step 3

3 of 8

b) $y=-3cos x+5$

(1) reflect over $x$-axis

(2) vertically stretch by a factor of 3 (multiply amplitude by 3)

(3) vertically translate upwards by 5

Exercise scan

Step 4

4 of 8

c) $y=-sin(6x)+4$

(1) reflect over $x-$axis

(2) horizontally compress by a factor of 6 (divide period by 6)

(3) vertically translate upwards by 4

Exercise scan

Step 5

5 of 8

d) $y=4cos(2x)-3$

(1) horizontally compress by a factor of 2 (divide period by 2)

(2) vertically stretch by a factor of 4 (multiply amplitude by 4)

(3) vertically translate downwards by 3

Exercise scan

Step 6

6 of 8

e) $y=dfrac{1}{2}cos(3x-120^circ)$

(1) vertically compress by a factor of 2 (multiply amplitude by 0.5)

(2) horizontally compress by a factor 3 (divide period by 3)

(3) horizontally translate to the right by $120^circ$

Exercise scan

Step 7

7 of 8

f) $y=-8sinleft[ dfrac{1}{2}(x+50^circ)right]-9$

(1) reflect over x-axis and horizontally stretch by a factor of 2

(2) horizontally translate by $50^circ$

(3) vertically stretch by a factor of 8 (multiply amplitude by 8)

(4) vertically translate downwards by 9

Exercise scan

Result

8 of 8

See explanation inside.

Exercise 8

Step 1

1 of 7

[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| hfill \
{text{period = }}T=frac{{{{360}^ circ }}}{|k|} hfill \
{text{equation of axis:}}{text{ }},,,y = c hfill \
hfill \
{text{If the function starts at }}x = {x_0}{text{ and goes through}},n{text{ cycles,}} hfill \
hfill \
{text{domain: }}left{ {x in {mathbf{R}};|;{text{ }}{x_0} leq x leq {x_0} + nT} right} hfill \
hfill \
{text{range: }}left{ {y in {mathbf{R}};|;{text{ }}c – |A| leq y leq c + |A|} right} hfill \
hfill \
hfill \
end{gathered} ]

Step 2

2 of 7

To determine the graphing WINDOW settings, we must know the domain and range. For one complete cycle, the domain is 1 period starting from the initial point and range must be within the WINDOW settings.

We will arbitrarily set the origin as the initial point.

Step 3

3 of 7

a) $k(x)=sin(2x)+6$

$T=dfrac{360^circ}{2}=180^circ$

$c-|A|=6-1=5$

$c+|A|=6+1=7$

Therefore, the settings must be

$x_{min}leq 0$ , $x_{max}geq 180^circ$

$y_{min}leq 5$ , $y_{max}geq 7$

Step 4

4 of 7

b) $j(x)=-5sinleft(dfrac{1}{2}xright)+20$

$T=dfrac{360^circ}{1/2}=720^circ$

$c-|A|=20-5=15$

$c+|A|=20+5=25$

Therefore, the settings must be

$x_{min}leq 0$ , $x_{max}geq 720^circ$

$y_{min}leq 15$ , $y_{max}geq 25$

Step 5

5 of 7

c) $j(x)=7cosleft(90(x-1^circ)right)+82$

$T=dfrac{360^circ}{1/2}=720^circ$

$c-|A|=82-7=75$

$c+|A|=82+7=89$

Therefore, the settings must be

$x_{min}leq 0$ , $x_{max}geq 720^circ$

$y_{min}leq 15$ , $y_{max}geq 25$

Step 6

6 of 7

d) $f(x)=0.5sin(360x+72^circ)=17$

$$
T=dfrac{360}{360}=1
$$

$c-|A|=-27-0.5=-27.5$

$c+|A|=-27+0.5=-26.5$

Therefore, the settings must be

$x_{min}leq 0$ , $x_{max}geq 1^circ$

$y_{min}leq -27.5$ , $y_{max}geq 26.5$

Result

7 of 7

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|}
hline
& Settings \ hline
a) & begin{tabular}[c]{@{}l@{}}$x_{min}leq 0^circ$ , $x_{max}geq 180^circ$\ $y_{min}leq 5$ , $y_{max}geq 7$end{tabular} \ hline
b) & begin{tabular}[c]{@{}l@{}}$x_{min}leq 0^circ$ , $x_{max}geq 720^circ$\ $y_{min}leq 15$ , $y_{max}geq 25$end{tabular} \ hline
c) & begin{tabular}[c]{@{}l@{}}$x_{min}leq 0^circ$ , $x_{max}geq 720^circ$\ $y_{min}leq 15$ , $y_{max}geq 25$end{tabular} \ hline
d) & begin{tabular}[c]{@{}l@{}}$x_{min}leq 0^circ$ , $x_{max}geq 1^circ$\ $y_{min}leq -27.5$ , $y_{max}geq 26.5$end{tabular} \ hline
end{tabular}
end{table}

Exercise 9

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| hfill \
{text{period = }}T=frac{{{{360}^ circ }}}{|k|} hfill \
{text{equation of axis:}}{text{ }},,,y = c hfill \
hfill \
{text{If the function starts at }}x = {x_0}{text{ and goes through}},n{text{ cycles,}} hfill \
hfill \
{text{domain: }}left{ {x in {mathbf{R}};|;{text{ }}{x_0} leq x leq {x_0} + nT} right} hfill \
hfill \
{text{range: }}left{ {y in {mathbf{R}};|;{text{ }}c – |A| leq y leq c + |A|} right} hfill \
hfill \
hfill \
end{gathered} ]

Step 2

2 of 3

$P(t)=-20cos(300t)^circ+100$

a) The period is $T=dfrac{360}{300}=1.2$ s

This represents the time it takes for one breathe cycle.

b)

$c-|A|=100-20=80$

$c+|A|=100+20=120$

range: ${ P(t) in bold{R};|;80leq P(t)leq 120}$

The range represents the set of all possible values of blood pressure within the normal range.

Result

3 of 3

a) $1.2$ s

b) ${ P(t) in bold{R};|;80leq P(t)leq 120}$

Exercise 10

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| hfill \
{text{period = }}T=frac{{{{360}^ circ }}}{|k|} hfill \
{text{equation of axis:}}{text{ }},,,y = c hfill \
hfill \
{text{If the function starts at }}x = {x_0}{text{ and goes through}},n{text{ cycles,}} hfill \
hfill \
{text{domain: }}left{ {x in {mathbf{R}};|;{text{ }}{x_0} leq x leq {x_0} + nT} right} hfill \
hfill \
{text{range: }}left{ {y in {mathbf{R}};|;{text{ }}c – |A| leq y leq c + |A|} right} hfill \
hfill \
hfill \
end{gathered} ]

Step 2

2 of 3

Given that $y_{min}=-1$ and $y_{max}=7$

The amplitude can be calculated as

$A=dfrac{|7-(-1)|}{2}=4$

The equation of axis is

$y=dfrac{-1+7}{2}implies y=3$

The period is $T=720^circ implies k=dfrac{360^circ}{720^circ}=0.5$

a) The equation of sine function is $y=4sin(0.5x)+3$

b) The equation of cosine function is $y=4cos(0.5x)+3$

Result

3 of 3

a) $y=4sin(0.5x)+3$

b) $y=4cos(0.5x)+3$

Exercise 11

Step 1

1 of 4

begin{table}[]
begin{tabular}{|l|l|}
hline
Transformation & Description \ hline
$y=f(x)+c$ & begin{tabular}[c]{@{}l@{}}vertical translation of\ $c$ units upwardend{tabular} \ hline
$y=f(x+d)$ & begin{tabular}[c]{@{}l@{}}horizontal translation of $d$ units\ to the leftend{tabular} \ hline
$y=acdot f(x)$ & begin{tabular}[c]{@{}l@{}}vertical stretching, multiplies\ the amplitude by $a$end{tabular} \ hline
$y=f(kx)$ & begin{tabular}[c]{@{}l@{}}horizontal compression, divides\ the period by $k$end{tabular} \ hline
$y=-f(x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in\ the $x$-axisend{tabular} \ hline
$y=f(-x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in \ the $y$-axisend{tabular} \ hline
end{tabular}
end{table}

Step 2

2 of 4

$f(x)=-dfrac{1}{2}cos(120x)+30$

(1) Start by graphing $y=cos x$. It should be a sinusoidal graph passing through $(0,1)$ with period of $360^circ$, amplitude of 1, and axis at $y=0$.

(2)reflect the graph over $x-$axis

(3) Horizontally compress by a factor of 120, that is, divide the period by 120.

(4) Vertically compress by a factor of 2, that is, divide the amplitude by 2.

(5) Vertically translate upwards by 30

Step 3

3 of 4

Result

4 of 4

See answers inside.

Exercise 12

Step 1

1 of 5

begin{table}[]
begin{tabular}{|l|l|}
hline
Transformation & Description \ hline
$y=f(x)+c$ & begin{tabular}[c]{@{}l@{}}vertical translation of\ $c$ units upwardend{tabular} \ hline
$y=f(x+d)$ & begin{tabular}[c]{@{}l@{}}horizontal translation of $d$ units\ to the leftend{tabular} \ hline
$y=acdot f(x)$ & begin{tabular}[c]{@{}l@{}}vertical stretching, multiplies\ the amplitude by $a$end{tabular} \ hline
$y=f(kx)$ & begin{tabular}[c]{@{}l@{}}horizontal compression, divides\ the period by $k$end{tabular} \ hline
$y=-f(x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in\ the $x$-axisend{tabular} \ hline
$y=f(-x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in \ the $y$-axisend{tabular} \ hline
end{tabular}
end{table}

Step 2

2 of 5

When the graph of $y=sin x$ and $y=cos x$ are subjected to horizontal compression of 0.5, that means the period is reduced to half. Thus, we get $y=sin 2x$ and $y=cos 2x$

Step 3

3 of 5

Step 4

4 of 5

From the graph, we see that the graphs can be superimposed by horizontally translating $y=cos 2x$ by 45$^circ$ to the right or translating $y=sin 2x$ by $45^circ$ to the left.

Result

5 of 5

See answers inside.

Exercise 13

Step 1

1 of 7

a) A trigonometric function is reasonable because the number of hours of daylight is a periodic quantity that completes a cycle in 1 year. It can also model the occurrence of maximum and minimum daylight hours that occur within a year.

Step 2

2 of 7

When the graph of $y=sin x$ and $y=cos x$ are subjected to horizontal compression of 0.5, that means the period is reduced to half. Thus, we get $y=sin 2x$ and $y=cos 2x$

Step 3

3 of 7

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|c|c|}
hline
multicolumn{1}{|l|}{Month} & multicolumn{1}{l|}{Number of Days} \ hline
January & 31 \ hline
February & 28 \ hline
March & 31 \ hline
April & 30 \ hline
May & 31 \ hline
June & 30 \ hline
July & 31 \ hline
August & 31 \ hline
September & 30 \ hline
October & 31 \ hline
November & 30 \ hline
December & 31 \ hline
end{tabular}
end{table}

Step 4

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b) From the table, March 21 is the $31+28+31-10=80$th day of the year. Substituting $t=80$ to the equation

$D(t)=4sinleft[dfrac{360}{365}(t-80)right]^circ+12$

$D(80)=4sinleft[dfrac{360}{365}(80-80)right]^circ+12=12$

September 21 is the $365-(31+30+31+9)=264$th day of the year.

$D(264)=4sinleft[dfrac{360}{365}(264-80)right]^circ+12=11.89$

This days correspond to the spring and fall equinox (equal day and night).

Step 5

5 of 7

c) June 21 is the $172$ day of the year,

$D(172)=4sinleft[dfrac{360}{365}(172-80)right]^circ+12=16$

This corresponds to the longest day or summer solstice.

December 21 is the 355th day of the year,

$D(355)=4sinleft[dfrac{360}{365}(355-80)right]^circ+12=8$

This corresponds to the shortest day or winter solstice.

Step 6

6 of 7

d) The 12 serves as the axis which is the halfway between maximum and minimum hours of daylight. This makes sense as half of 24 hours (full day) is 12.

Result

7 of 7

a) see explanation inside

b) 12 hrs, 11.89 hrs

c) 16 hrs, 8 hrs

d) see explanation inside

All Solutions

Section 6-5: Using Transformations to Sketch the Graphs of Sinusoidal Functions

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