Wize High School Algebra II Textbook (Common Core) > Trigonometric (Sinusoidal) Graphs

Transformations of Sinusoidal Functions

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Transformations of Sinusoidal Functions (Pt 1)

Let y=af(b(x−h))+ky=af(b(x-h))+ky=af(b(x−h))+k where f(x)f(x)f(x) is a trigonometric function. Then, 


HorizontalVerticalb<0‾:a<0‾:Reflection about theReflection about they-axisx-axis∣b∣>1‾:∣a∣>1‾:Compresses the period  1b  unitsExpansion of ’a’ units∣b∣<1‾:∣a∣<1‾:Expands the period  1b  unitsCompression of ’a’ unitsh>0‾:k>0‾:Phase shift ’h’ units rightDisplacement ’k units uph<0‾:k<0‾:Phase shift ’h’ units leftDisplacement ’k’ units down\begin{array}{l c c l}
\textbf{Horizontal}&&&\textbf{Vertical}\\\\
\underline{b<0}:&&&\underline{a<0}:\\
\text{Reflection about the}&&&\text{Reflection about the}\\
\text{y-axis}&&&\text{x-axis}\\\\\\
\underline{|b|>1}:&&&\underline{|a|>1}:\\
\text{Compresses the period}~~\displaystyle\frac{1}{b}~~\text{units}&&&\text{Expansion of 'a' units}\\\\\\
\underline{|b|<1}:&&&\underline{|a|<1}:\\
\text{Expands the period}~~\displaystyle\frac{1}{b}~~\text{units}&&&\text{Compression of 'a' units}\\\\\\
\underline{h>0}:&&&\underline{k>0}:\\
\text{Phase shift 'h' units right}&&&\text{Displacement 'k
 units up}\\\\\\
\underline{h<0}:&&&\underline{k<0}:\\
\text{Phase shift 'h' units left}&&&\text{Displacement 'k' units down}
\end{array}Horizontalb<0​:Reflection about they-axis∣b∣>1​:Compresses the period  b1​  units∣b∣<1​:Expands the period  b1​  unitsh>0​:Phase shift ’h’ units righth<0​:Phase shift ’h’ units left​​​Verticala<0​:Reflection about thex-axis∣a∣>1​:Expansion of ’a’ units∣a∣<1​:Compression of ’a’ unitsk>0​:Displacement ’k units upk<0​:Displacement ’k’ units down​


Wize Tip
The value 2π∣b∣\colorThree{\displaystyle\frac{2\pi}{|b|}}∣b∣2π​ is the period. 

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How to Graph the Transformation of a Sine or Cosine Function
Step 1. 
Find the period. 

P=2πbP=\displaystyle\frac{2\pi}{b}P=b2π​

Count 8 spaces from the origin (left and/or right).
Label the period.
Label the rest of the graph.


Step 2.
Find the amplitude 'aaa 


Step 3. 
Find the vertical displacement 'kkk'.


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Step 4.
Identify:

 Y−Maximum=a+kY−Average=kY−Minimum=−a+k\begin{array}{lcl}
Y-\text{Maximum}&=&a+k\\\\
Y-\text{Average}&=&k\\\\
Y-\text{Minimum}&=&-a+k
\end{array}Y−MaximumY−AverageY−Minimum​===​a+kk−a+k​


Step 5.  

Identify the phase shift and determine how many units to the left or right the function must move.

n=Phase Shift(Period÷8)n=\displaystyle\frac{\text{Phase Shift}}{(\text{Period}\div8)}n=(Period÷8)Phase Shift​


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Step 6.

Graph.

If sin⁡θ\sin{\theta}sinθ:

Begin on a Y-Average\color{Red}{\text{Y-Average}}Y-Average. 
Count 2 units right and plot a Y-Maximum\color{red}{\text{Y-Maximum}}Y-Maximum.
Count 2 units right and plot a Y-Average\color{red}\text{Y-Average}Y-Average.
Count 2 units right and plot a Y-Minimum\color{red}\text{Y-Minimum}Y-Minimum.
Count 2 units right and plot a Y-Average\color{red}\text{Y-Average}Y-Average.
Continue this pattern in the right/left direction. 

If cos⁡θ\cos{\theta}cosθ:

Begin on a Y-Maximum\color{red}\text{Y-Maximum}Y-Maximum.
Count 2 units right and plot a Y-Average\color{red}\text{Y-Average}Y-Average.
Count 2 units right and plot a Y-Minimum\color{red}\text{Y-Minimum}Y-Minimum.
Count 2 units right and plot a Y-Average\color{red}\text{Y-Average}Y-Average.
Count 2 units right and plot a Y-Maximum\color{red}\text{Y-Maximum}Y-Maximum.
Continue this pattern in the right/left direction. 
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Example 1

Let's look at y=2sin⁡(2(θ−π4))+1y=2\sin(2(\theta-\frac{\pi}{4}))+1y=2sin(2(θ−4π​))+1.

Step 1. 
Find the period. 

P=2π2=πP=\displaystyle\frac{2\pi}{2}=\piP=22π​=π

Count 8 spaces from the origin (left & right).
Label the period.
Label the rest of the graph. 

  

Step 2.
Find the amplitude.

a=2a=2a=2


Step 3. 
Find the vertical displacement. 

k=1k=1k=1

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Step 4.
Identify the Y−Maximum, Y−average, Y−Minimum\colorThree{Y-\text{Maximum},~Y-\text{average},~Y-\text{Minimum}}Y−Maximum, Y−average, Y−Minimum.

Y-Min=−1\text{Y-Min}=-1Y-Min=−1

Y-Avg=1\text{Y-Avg}=1Y-Avg=1

Y-Max=3\text{Y-Max}=3Y-Max=3


Step 5.  
Identify the phase shift and determine how many units to the left or right the function must move.

The phase shift is π4\displaystyle\frac{\pi}{4}4π​ units right. 

So, how many cartesian coordinates right is this? 

h=π4(π÷8)=2h=\displaystyle\frac{\frac{\pi}{4}}{(\pi\div{8})}=2h=(π÷8)4π​​=2

This means that π4\displaystyle\frac{\pi}{4}4π​ is 2 units right from the origin.

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Step 6.

Since we are graphing sin⁡θ\sin{\theta}sinθ , then we begin on an Y-Average\color{red}\text{Y-Average}Y-Average of (π4,1)\Bigg(\displaystyle\frac{\pi}{4},1\Bigg)(4π​,1)

Count 2 spaces to the right and plot the Y-Maximum\color{red}\text{Y-Maximum}Y-Maximum at (π2,3)\Bigg(\displaystyle\frac{\pi}{2},3\Bigg)(2π​,3)

Count 2 spaces to the right and plot the Y-Average\color{red}\text{Y-Average}Y-Average of (3π4,1)\Bigg(\displaystyle\frac{3\pi}{4},1\Bigg)(43π​,1)

Count 2 spaces to the right and plot the Y-Minimum\color{red}\text{Y-Minimum}Y-Minimum of (π,−1)(\pi,-1)(π,−1)


Count 2 spaces to the right and plot the Y-Average\color{red}\text{Y-Average}Y-Average of (5π4,1)\Bigg(\displaystyle\frac{5\pi}{4},1\Bigg)(45π​,1)

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Continue on with this pattern in both directions, 

  

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Example: Transformations of Sinusoidal Functions
Graph y=−3sin⁡(2(θ+π3))+1y=-3\sin\Bigg(2\Big(\theta+\displaystyle\frac{\pi}{3}\Big)\Bigg)+1y=−3sin(2(θ+3π​))+1.


Step 1. 
Find the period. 

P=2π2=πP=\displaystyle\frac{2\pi}{2}=\piP=22π​=π

Count 8 spaces from the origin (left & right).
Label the period.
Label the rest of the graph.
  

Step 2.
Find the amplitude.

a=3a=3a=3

Step 3. 
Find the vertical displacement. 

k=1k=1k=1

Step 4.
Identify the Y−Maximum, Y−average, Y−Minimum\colorThree{Y-\text{Maximum},~Y-\text{average},~Y-\text{Minimum}}Y−Maximum, Y−average, Y−Minimum.

Y-Min=−(−3)+1=4\text{Y-Min}=-(-3)+1=4Y-Min=−(−3)+1=4

Y-Avg=1\text{Y-Avg}=1Y-Avg=1

Y-Max=−3+1=−2\text{Y-Max}=-3+1=-2Y-Max=−3+1=−2

Step 5.  
Identify the phase shift and determine how many units to the left or right the function must move.

The phase shift is π3\displaystyle\frac{\pi}{3}3π​ units left. 

Therefore, 

h=π3(π÷8)=83h=\displaystyle\frac{\frac{\pi}{3}}{(\pi\div{8})}=\displaystyle\frac{8}{3}h=(π÷8)3π​​=38​

This means that π3\displaystyle\frac{\pi}{3}3π​ is 83\displaystyle\frac{8}{3}38​ units left of the origin.

Step 6.

Since we are graphing sin⁡θ\sin{\theta}sinθ , then we begin on an Y-Average\color{red}\text{Y-Average}Y-Average of (−π3,1)\Bigg(-\displaystyle\frac{\pi}{3},1\Bigg)(−3π​,1)

Count 2 spaces to the right and plot the Y-Maximum\color{red}\text{Y-Maximum}Y-Maximum at (−π12,−2)\Bigg(-\displaystyle\frac{\pi}{12},-2\Bigg)(−12π​,−2)

Count 2 spaces to the right and plot the Y-Average\color{red}\text{Y-Average}Y-Average of (π6,1)\Bigg(\displaystyle\frac{\pi}{6},1\Bigg)(6π​,1)

Count 2 spaces to the right and plot the Y-Minimum\color{red}\text{Y-Minimum}Y-Minimum of (5π6,4)\Bigg(\displaystyle\frac{5\pi}{6},4\Bigg)(65π​,4)

Count 2 spaces to the right and plot the Y-Average\color{red}\text{Y-Average}Y-Average of (2π3,1)\Bigg(\displaystyle\frac{2\pi}{3},1\Bigg)(32π​,1)

Continue on with this pattern in both directions.

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Transformations of Sinusoidal Functions (Pt 2)

How to Graph the Transformation of a Tangent Function y=atan⁡(b(θ−h)+k\colorOne{y=a\tan{(b(\theta-h)+k}}y=atan(b(θ−h)+k 

Step 1. 
Find the period. 

P=πbP=\displaystyle\frac{\pi}{b}P=bπ​

Count 4 spaces from the origin (left & right).
Label the period.
Label the rest of the graph.  

Step 2. 
Identify the phase shift and calculate how many units left/right the shift is.

h=Phase Shift(Period÷4)h=\displaystyle\frac{\text{Phase Shift}}{(\text{Period}\div4)}h=(Period÷4)Phase Shift​

Step 3.
Identify the x-intercepts.

(0,0)±πn, n∈W  →  (0b+h ,0)±πbn, n∈W(0,0)\pm{}\pi{}n,~n\in\mathbb{W}~~{\color{red}\rightarrow}~~\Bigg(\displaystyle\frac{0}{\colorThree{b}}+{\colorThree{h}}~,0\Bigg)\pm{}\displaystyle\frac{\pi}{\colorThree{b}}{}n,~n\in\mathbb{W}(0,0)±πn, n∈W  →  (b0​+h ,0)±bπ​n, n∈W

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Step 4.
Identify the vertical asymptotes. 

(π2,0)±πn, n∈W  →  ((π2÷b)+h ,0)±πbn, n∈W\Bigg(\displaystyle\frac{\pi}{2},0\Bigg)\pm{}\pi{}n,~n\in\mathbb{W}~~{\color{red}\rightarrow}~~\Bigg(\Big(\displaystyle\frac{\pi}{2}\div{\colorThree{b}}\Big)+{\colorThree{h}}~,0\Bigg)\pm{}\displaystyle\frac{\pi}{\colorThree{b}}{}n,~n\in\mathbb{W}(2π​,0)±πn, n∈W  →  ((2π​÷b)+h ,0)±bπ​n, n∈W

Step 5. 
The points (−π4,−1)  &  (π4,1) ±πn,  n∈W\Bigg(\displaystyle-\frac{\pi}{4},-1\Bigg)~~\&~~\Bigg(\displaystyle\frac{\pi}{4},1\Bigg)~\pm{}\pi{}n,~~n\in\mathbb{W}(−4π​,−1)  &  (4π​,1) ±πn,  n∈W are on the graph of y=tan⁡θy=\tan{\theta}y=tanθ.

Find the points on the transformed function.

((−π4÷b)+h,−1a+k)  &  ((π4÷b+h,1a+k) ±πbn,  n∈W\Bigg(\Big(\displaystyle-\frac{\pi}{4}\div{\colorThree{b}}\Big)+\colorThree{h},-1\colorThree{a}+\colorThree{k}\Bigg)~~\&~~\Bigg(\Big(\displaystyle\frac{\pi}{4}\div{\colorThree{b}}+\colorThree{h},1\colorThree{a}+\colorThree{k}\Bigg)~\pm{}\displaystyle\frac{\pi}{\colorThree{b}}n,~~n\in\mathbb{W}((−4π​÷b)+h,−1a+k)  &  ((4π​÷b+h,1a+k) ±bπ​n,  n∈W

Step 6.

Graph.

Begin with an x-intercept\color{red}\text{x-intercept}x-intercept 
Count 2 units right and draw a vertical asymptote\color{red}\text{vertical asymptote}vertical asymptote
Count 2 units right and draw an x-intercept\color{red}\text{x-intercept}x-intercept
Count 2 units right and draw a vertical asymptote\color{red}\text{vertical asymptote}vertical asymptote
Count 2 units right and draw an x-intercept\color{red}\text{x-intercept}x-intercept
Continue this pattern right/left. 

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Example 

Let's look at y=2tan⁡2(θ−π16)y=2\tan{2(\theta-\frac{\pi}{16})}y=2tan2(θ−16π​).

Step 1. 

P=π2P=\displaystyle\frac{\pi}{2}P=2π​

Count 4 spaces from the origin (left & right).
Label the period.
Label the rest of the graph. 
  
Step 2. 

h=π16(π2÷4)=0.5h=\displaystyle\frac{\frac{\pi}{16}}{(\frac{\pi}{2}\div4)}=0.5h=(2π​÷4)16π​​=0.5

Therefore, π16\displaystyle\frac{\pi}{16}16π​ is 0.5 units right. 

Step 3.

X-intercepts: 

(02+π16, 0)±π2n, n∈W  →  (π16,0)±π2n, n∈W\Bigg(\displaystyle\frac{0}{2}+\displaystyle\frac{\pi}{16},~0\Bigg)\pm{}\displaystyle\frac{\pi}{2}n,~n\in\mathbb{W}~~{\color{red}\rightarrow}~~\Bigg(\displaystyle\frac{\pi}{16},0\Bigg)\pm{}\displaystyle\frac{\pi}{2}n,~n\in\mathbb{W}(20​+16π​, 0)±2π​n, n∈W  →  (16π​,0)±2π​n, n∈W

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Step 4.

Vertical asymptotes:

(π2÷2+π16)±π2n, n∈W  →  5π16±π2n, n∈W\Bigg(\displaystyle\frac{\pi}{2}\div{2}+\displaystyle\frac{\pi}{16}\Bigg)\pm{}\displaystyle\frac{\pi}{2}n,~n\in\mathbb{W}~~{\color{red}\rightarrow}~~\displaystyle\frac{5\pi}{16}\pm{}\displaystyle\frac{\pi}{2}n,~n\in\mathbb{W}(2π​÷2+16π​)±2π​n, n∈W  →  165π​±2π​n, n∈W

Step 5.

(−π4,−1)  &  (π4,1) ±πn,  n∈W   →   (−π16,−2)  &  (3π16,2) ±π2n,  n∈W\Bigg(\displaystyle-\frac{\pi}{4},-1\Bigg)~~\&~~\Bigg(\displaystyle\frac{\pi}{4},1\Bigg)~\pm{}\pi{}n,~~n\in\mathbb{W}~~~{\color{red}\rightarrow}~~~\Bigg(\displaystyle-\frac{\pi}{16},-2\Bigg)~~\&~~\Bigg(\displaystyle\frac{3\pi}{16},2\Bigg)~\pm{}\frac{\pi}{2}{}n,~~n\in\mathbb{W}(−4π​,−1)  &  (4π​,1) ±πn,  n∈W   →   (−16π​,−2)  &  (163π​,2) ±2π​n,  n∈W

Step 6.

Graph.

Begin with an x-intercept\color{red}\text{x-intercept}x-intercept at (π16,0)\Bigg(\displaystyle\frac{\pi}{16},0\Bigg)(16π​,0)
Count 2 units right and draw a vertical asymptote\color{red}\text{vertical asymptote}vertical asymptote at x=5π16x=\displaystyle\frac{5\pi}{16}x=165π​
Count 2 units right and draw an x-intercept\color{red}\text{x-intercept}x-intercept at (9π16,0)\Bigg(\displaystyle\frac{9\pi}{16},0\Bigg)(169π​,0)
Count 2 units right and draw a vertical asymptote\color{red}\text{vertical asymptote}vertical asymptote at x=13π16x=\displaystyle\frac{13\pi}{16}x=1613π​
Count 2 units right and draw an x-intercept\color{red}\text{x-intercept}x-intercept (17π16,0)\Bigg(\displaystyle\frac{17\pi}{16},0\Bigg)(1617π​,0)
Count 2 units right and draw a vertical asymptote\color{red}\text{vertical asymptote}vertical asymptote at x=21π16x=\displaystyle\frac{21\pi}{16}x=1621π​
Continue this pattern right/left.

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Plot the points from Step 5 and connect the dots, trending towards the asymptotes.

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Example: Transformations of Sinusoidal Functions
Graph y=2tan⁡(3(θ−π12))y=2\tan\Bigg(3\Big(\theta-\displaystyle\frac{\pi}{12}\Big)\Bigg)y=2tan(3(θ−12π​)).


Step 1. 

P=π3P=\displaystyle\frac{\pi}{3}P=3π​

Count 4 spaces from the origin (left & right).
Label the period.
Label the rest of the graph.  
  

Step 2. 

h=π12π3÷4=1h=\displaystyle\frac{\frac{\pi}{12}}{\frac{\pi}{3}\div4}=1h=3π​÷412π​​=1 unit to the right. 

Step 3.

(03+π12, 0)±π3n, n∈W  →  (π12,0)±π3n, n∈W\Bigg(\displaystyle\frac{0}{3}+\displaystyle\frac{\pi}{12},~0\Bigg)\pm{}\displaystyle\frac{\pi}{3}n,~n\in\mathbb{W}~~{\color{red}\rightarrow}~~\Bigg(\displaystyle\frac{\pi}{12},0\Bigg)\pm{}\displaystyle\frac{\pi}{3}n,~n\in\mathbb{W}(30​+12π​, 0)±3π​n, n∈W  →  (12π​,0)±3π​n, n∈W

Step 4.

(π2÷3+π12)±π3n, n∈W  →  π4±π3n, n∈W\Bigg(\displaystyle\frac{\pi}{2}\div{3}+\displaystyle\frac{\pi}{12}\Bigg)\pm{}\displaystyle\frac{\pi}{3}n,~n\in\mathbb{W}~~{\color{red}\rightarrow}~~\displaystyle\frac{\pi}{4}\pm{}\displaystyle\frac{\pi}{3}n,~n\in\mathbb{W}(2π​÷3+12π​)±3π​n, n∈W  →  4π​±3π​n, n∈W

Step 5. 

(−π4,−1)  &  (π4,1) ±πn,  n∈W   →   (0,−2)  &  (π6,2) ±π3n,  n∈W\Bigg(\displaystyle-\frac{\pi}{4},-1\Bigg)~~\&~~\Bigg(\displaystyle\frac{\pi}{4},1\Bigg)~\pm{}\pi{}n,~~n\in\mathbb{W}~~~{\color{red}\rightarrow}~~~\Big(0,-2\Big)~~\&~~\Bigg(\displaystyle\frac{\pi}{6},2\Bigg)~\pm{}\frac{\pi}{3}{}n,~~n\in\mathbb{W}(−4π​,−1)  &  (4π​,1) ±πn,  n∈W   →   (0,−2)  &  (6π​,2) ±3π​n,  n∈W

Step 6.

Graph.

Begin with an x-intercept\color{red}\text{x-intercept}x-intercept at (π12,0)\Bigg(\displaystyle\frac{\pi}{12},0\Bigg)(12π​,0)
Count 2 units right and draw a vertical asymptote\color{red}\text{vertical asymptote}vertical asymptote at x=π4x=\displaystyle\frac{\pi}{4}x=4π​
Count 2 units right and draw an x-intercept\color{red}\text{x-intercept}x-intercept at (5π12,0)\Bigg(\displaystyle\frac{5\pi}{12},0\Bigg)(125π​,0)
Count 2 units right and draw a vertical asymptote\color{red}\text{vertical asymptote}vertical asymptote at x=7π12x=\displaystyle\frac{7\pi}{12}x=127π​
Count 2 units right and draw an x-intercept\color{red}\text{x-intercept}x-intercept (3π4,0)\Bigg(\displaystyle\frac{3\pi}{4},0\Bigg)(43π​,0)
Count 2 units right and draw a vertical asymptote\color{red}\text{vertical asymptote}vertical asymptote at x=11π12x=\displaystyle\frac{11\pi}{12}x=1211π​
Continue this pattern right/left.
Plot the points from Step 5 and connect the dots, trending towards the asymptotes. 
  

Practice: Transformations of Sinusoidal Functions
The following is a table of values describing a sinusoidal relationship between xxx and y.y.y.

xπ12π45π127π123π4y0−5050\begin{array}{|c|c|c|c|c|c|}\hline\\
\textbf{x}&\displaystyle\frac{\pi}{12}&\displaystyle\frac{\pi}{4}&\displaystyle\frac{5\pi}{12}&\displaystyle\frac{7\pi}{12}&\displaystyle\frac{3\pi}{4}\\\\\hline\\
\textbf{y}&0&-5&0&5&0\\\\\hline
\end{array}xy​12π​0​4π​−5​125π​0​127π​5​43π​0​​

What function best describes the tables of values? 

y=−5sin⁡(3(x−π12))y=-5\sin{\Bigg(3\Big(x-\displaystyle\frac{\pi}{12}\Big)\Bigg)}y=−5sin(3(x−12π​)) 

y=5sin⁡(3(x−π12))y=5\sin{\Bigg(3\Big(x-\displaystyle\frac{\pi}{12}\Big)\Bigg)}y=5sin(3(x−12π​)) 

y=−5sin⁡(3(x+π12))y=-5\sin{\Bigg(3\Big(x+\displaystyle\frac{\pi}{12}\Big)\Bigg)}y=−5sin(3(x+12π​)) 

 y=5sin⁡(3(x+π12))y=5\sin{\Bigg(3\Big(x+\displaystyle\frac{\pi}{12}\Big)\Bigg)}y=5sin(3(x+12π​)) 

I don't know

Practice: Transformations of Sinusoidal Functions
Let y=cos⁡θy=\cos{\theta}y=cosθ undergo the following transformations:
A vertical compression by a factor of 12\displaystyle\frac{1}{2}21​
A horizontal compression by a factor of 14\displaystyle\frac{1}{4}41​
A phase shift/horizontal translation of 3π16\displaystyle\frac{3\pi}{16}163π​ units left
Which of the following graphs best displays the graph of the transformed function?

Practice: Transformations of Sinusoidal Functions
If (π4,3)\Bigg(\displaystyle\frac{\pi}{4}, 3\Bigg)(4π​,3) is on the function y=2cos⁡2(θ−π3)−3y=2\cos{2\Big(\theta-\displaystyle\frac{\pi}{3}\Big)}-3y=2cos2(θ−3π​)−3, what point must be on the function y=cos⁡θy=\cos{\theta}y=cosθ?

Answer

I don't know

Extra Practice

Transformations of Sinusoidal Functions

Practice: Transformations of Sinusoidal Functions
If (π4,3)\Bigg(\displaystyle\frac{\pi}{4}, 3\Bigg)(4π​,3) is on the function y=2cos⁡2(θ−π3)−3y=2\cos{2\Big(\theta-\displaystyle\frac{\pi}{3}\Big)}-3y=2cos2(θ−3π​)−3, what point must be on the function y=cos⁡θy=\cos{\theta}y=cosθ?

Transformations of Sinusoidal Functions

Practice: Transformations of Sinusoidal Functions
Let y=cos⁡θy=\cos{\theta}y=cosθ undergo the following transformations:
A vertical compression by a factor of 12\displaystyle\frac{1}{2}21​

Transformations of Sinusoidal Functions

Practice: Transformations of Sinusoidal Functions
The following is a table of values describing a sinusoidal relationship between xxx and y.y.y.
xπ12π45π127π123π4y0−5050\begin{array}{|c|c|c|c|c|c|}\hline\\
\textbf{x}&\displaystyle\frac{\pi}{12}&\displaystyle\frac{\pi}{4}&\displaystyle\frac{5\pi}{12}&\displaystyle\frac{7\pi}{12}&\displaystyle\frac{3\pi}{4}\\\\\hline\\
\textbf{y}&0&-5&0&5&0\\\\\hline
\end{array}xy​12π​0​4π​−5​125π​0​127π​5​43π​0​​

Equivalent Sinusoidal Functions

Let y=3cos⁡4(x+π16)y=3\cos4\Big(x+\frac{\pi}{16}\Big)y=3cos4(x+16π​). 
Write an equivalent expression using sin⁡θ\sin\thetasinθ with the smallest possible phase shift and a>0a>0a>0.

Transformations of Sinusoidal Functions

Sketch a graph of the function y=2cos⁡2(θ−π8)y=2\cos{2\Big(\theta-\dfrac{\pi}{8}\Big)}y=2cos2(θ−8π​)

Transformations of Sinusoidal Functions

A sinusoidal function has the following properties:
Ymax=−1Y_{\text{max}}=-1Ymax​=−1
Yavg=1Y_{\text{avg}}=1Yavg​=1