Unit 3 · Topic 3.6

Sinusoidal Function Transformations

In f(θ) = asin (b(θ + c)) + d , |a| is the amplitude, 2π/(|b|) the period, c the phase shift (left by c ) and y = d the midline; cosine works the same way.

Practice 1: Procedural and Symbolic FluencyPractice 2: Multiple Representations

Part 1 · Hook

Why this matters

A music app draws a sound wave, and an engineer needs its formula to filter it: how tall, how fast, how far along. Every sinusoid you will ever meet is y=sin⁡θy = \sin\theta stretched, squeezed and slid. Four constants, aa, bb, cc and dd, record exactly how, and reading them correctly is the heart of the periodic modeling question on the exam.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. The greatest value of a sinusoidal function is 99 and the least is 11. What is its midline?

  1. y=5y = 5
  2. y=4y = 4
  3. y=8y = 8
  4. y=10y = 10
Show the answer

The midline is the average of the extremes: 9+12=5\frac{9 + 1}{2} = 5.

  • Correct: y=5y = 5:
  • y=4y = 4:
  • y=8y = 8:
  • y=10y = 10:

2. The graph of y=f(x−3)y = f(x - 3) is the graph of y=f(x)y = f(x)

  1. moved 3 units right
  2. moved 3 units left
  3. moved 3 units down
  4. stretched horizontally by 3
Show the answer

The input xx must be 3 larger to give the same output, so every point moves 3 units right.

  • Correct: moved 3 units right:
  • moved 3 units left:
  • moved 3 units down:
  • stretched horizontally by 3:

Part 4 · See it

See it first

Dashed gray: the graph of y = sin theta. Green: the graph of g(theta) = 2 sin(theta − pi/4) + 1 for theta from −pi/4 to 9 pi over 4. Compared with sine, g is moved right by pi/4 (the phase shift, shown by an arrow from the origin to (pi/4, 1)), stretched vertically so its amplitude is 2, and moved up so its midline is y = 1 (dashed). Its maximum 3 is at 3 pi over 4 and its minimum −1 at 7 pi over 4; its period is still 2 pi.
g(θ)g(\theta)=2sin⁡(θ−π4)+1= 2\sin\left(\theta - \frac{\pi}{4}\right) + 1 is the sine graph (dashed) moved right π4\frac{\pi}{4}, stretched to amplitude 2 and lifted to the midline y=1y = 1. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Multiply by aa: asin⁡θa\sin\thetaa vertical dilation: the amplitude becomes |a||a|, and a negative aa also reflects the graph across the midline
  2. Multiply the input by bb: sin⁡(bθ)\sin(b\theta)a horizontal dilation: one cycle now takes 2π|b|\frac{2\pi}{|b|}, so the period is 2π|b|\frac{2\pi}{|b|} and b=2πperiodb = \frac{2\pi}{\text{period}}
  3. Add cc inside, after factoring out bb: sin⁡(b(θ+c))\sin(b(\theta + c))a phase shift: the graph moves left cc units if c>0c > 0, right if c<0c < 0
  4. Add dd outside: asin⁡(b(θ+c))+da\sin(b(\theta + c)) + da vertical translation: the midline becomes y=dy = d

Part 6 · Key ideas

Key ideas

  • For f(θ)f(\theta)=asin⁡(b(θ+c))+d= a\sin(b(\theta + c)) + d or acos⁡(b(θ+c))+da\cos(b(\theta + c)) + d: amplitude |a||a|, period 2π|b|\frac{2\pi}{|b|}, phase shift −c-c (left by cc), midline y=dy = d.
  • Factor bb out before reading the shift: sin⁡(2θ+π)\sin(2\theta + \pi)=sin⁡(2(θ+π2))= \sin\left(2\left(\theta + \frac{\pi}{2}\right)\right) moves left π2\frac{\pi}{2}, not π\pi.
  • Building from a graph: dd from the midline, |a||a| from the amplitude, b=2πperiodb = \frac{2\pi}{\text{period}}, then cc from one landmark (a rising midline crossing for sine, a highest point for cosine).
  • Many formulas give the same graph: cos⁡θ=sin⁡(θ+π2)\cos\theta = \sin\left(\theta + \frac{\pi}{2}\right), and cc can change by any whole number of periods.

Part 7 · Misconception

A common mistake

The wrong idea: In f(θ)=sin⁡(2θ−π)f(\theta) = \sin(2\theta - \pi) the graph moves right π\pi, and the period is 2⋅2π=4π2 \cdot 2\pi = 4\pi.

What actually happens: Factor first: sin⁡(2θ−π)\sin(2\theta - \pi)=sin⁡(2(θ−π2))= \sin\left(2\left(\theta - \frac{\pi}{2}\right)\right), a shift right π2\frac{\pi}{2}. Multiplying the input by 2 makes each cycle shorter, so the period is 2π2=π\frac{2\pi}{2} = \pi.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

A sinusoidal function gg

π/2π3π/22π−3−2−112345θyg

1. Write a formula for g(θ)g(\theta).

Type an expression for g(θ)=g(\theta) =.

Show the answer

Midline d=4+(−2)2=1d = \frac{4 + (-2)}{2} = 1, amplitude 4−(−2)2=3\frac{4 - (-2)}{2} = 3, period 2π2\pi so b=1b = 1. The graph rises through its midline at θ=π2\theta = \frac{\pi}{2}, so g(θ)g(\theta)=3sin⁡(θ−π2)+1= 3\sin\left(\theta - \frac{\pi}{2}\right) + 1. Equivalent: −3cos⁡θ+1-3\cos\theta + 1.

  • Answer: g(θ)=g(\theta) = 3sin⁡(θ−π2)+13\,\sin\left(\theta - \frac{\pi}{2}\right) + 1

2. The function can be written g(θ)=3sin⁡(θ+c)+1g(\theta) = 3\sin(\theta + c) + 1. Which value of cc works?

  1. c=−π2c = -\frac{\pi}{2}
  2. c=π2c = \frac{\pi}{2}
  3. c=πc = \pi
  4. c=0c = 0
Show the answer

The sine pattern starts (rising midline crossing) at θ=−c\theta = -c. On the graph that is θ=π2\theta = \frac{\pi}{2}, so c=−π2c = -\frac{\pi}{2}.

  • Correct: c=−π2c = -\frac{\pi}{2}: Start of the cycle at θ=π2\theta = \frac{\pi}{2} means θ+c=0\theta + c = 0 there.
  • c=π2c = \frac{\pi}{2}: This moves the graph left; it would rise through the midline at −π2-\frac{\pi}{2} and fall through it at π2\frac{\pi}{2}.
  • c=πc = \pi: With c=πc = \pi the graph would have a lowest point at π2\frac{\pi}{2}, but gg is at its midline there.
  • c=0c = 0: With no shift the graph would be at its midline at θ=0\theta = 0, but g(0)=−2g(0) = -2.

3. Which is also a formula for gg?

  1. g(θ)=−3cos⁡θ+1g(\theta) = -3\cos\theta + 1
  2. g(θ)=3cos⁡θ+1g(\theta) = 3\cos\theta + 1
  3. g(θ)=−3sin⁡θ+1g(\theta) = -3\sin\theta + 1
  4. g(θ)g(\theta)=3cos⁡(θ−π2)+1= 3\cos\left(\theta - \frac{\pi}{2}\right) + 1
Show the answer

−3cos⁡θ+1-3\cos\theta + 1 is −3+1=−2-3 + 1 = -2 at θ=0\theta = 0 and 3+1=43 + 1 = 4 at θ=π\theta = \pi: a reflected cosine starts at a lowest point, as gg does.

  • Correct: g(θ)=−3cos⁡θ+1g(\theta) = -3\cos\theta + 1: A reflected cosine: lowest at 00, highest at π\pi.
  • g(θ)=3cos⁡θ+1g(\theta) = 3\cos\theta + 1: This has its highest point at θ=0\theta = 0; gg has its lowest point there.
  • g(θ)=−3sin⁡θ+1g(\theta) = -3\sin\theta + 1: This is 11 at θ=0\theta = 0, but g(0)=−2g(0) = -2.
  • g(θ)g(\theta)=3cos⁡(θ−π2)+1= 3\cos\left(\theta - \frac{\pi}{2}\right) + 1: cos⁡(θ−π2)=sin⁡θ\cos\left(\theta - \frac{\pi}{2}\right) = \sin\theta, so this is 11 at θ=0\theta = 0.

4. What is the average rate of change of gg on [0,π][0, \pi]?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

g(π)−g(0)π−0\frac{g(\pi) - g(0)}{\pi - 0}=4−(−2)π= \frac{4 - (-2)}{\pi}=6π= \frac{6}{\pi}, the slope of the secant line from the lowest point to the highest point.

  • Answer: 6π\frac{6}{\pi} ≈ 1.910

5. What is the period of f(θ)=5sin⁡(3θ)f(\theta) = 5\sin(3\theta)?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Period =2π|b|=2π3= \frac{2\pi}{|b|} = \frac{2\pi}{3}.

  • Answer: 2π3\frac{2\,\pi}{3} ≈ 2.094

6. How is the graph of f(θ)=sin⁡(2θ+π2)f(\theta) = \sin\left(2\theta + \frac{\pi}{2}\right) shifted, compared with y=sin⁡(2θ)y = \sin(2\theta)?

  1. Left π4\frac{\pi}{4}
  2. Left π2\frac{\pi}{2}
  3. Right π4\frac{\pi}{4}
  4. Right π2\frac{\pi}{2}
Show the answer

Factor out bb: 2θ+π2=2(θ+π4)2\theta + \frac{\pi}{2} = 2\left(\theta + \frac{\pi}{4}\right), so c=π4c = \frac{\pi}{4}: a shift left π4\frac{\pi}{4}.

  • Correct: Left π4\frac{\pi}{4}: After factoring out 2, the shift is π4\frac{\pi}{4} to the left.
  • Left π2\frac{\pi}{2}: This reads the shift before factoring out b=2b = 2.
  • Right π4\frac{\pi}{4}: A plus sign inside moves the graph left, not right.
  • Right π2\frac{\pi}{2}: Wrong direction, and bb was not factored out.

Part 9 · Summary

Summary

In f(θ)f(\theta)=asin⁡(b(θ+c))+d= a\sin(b(\theta + c)) + d, |a||a| is the amplitude, 2π|b|\frac{2\pi}{|b|} the period, cc the phase shift (left by cc) and y=dy = d the midline; cosine works the same way. To build a formula from a graph, read dd and |a||a|, get b=2πperiodb = \frac{2\pi}{\text{period}}, then find cc from a landmark point. Factor out bb before reading a shift.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

Question set for this topic