Mathematics Advanced • Year 11 • Module 2 • Lesson 12
Phase Shifts and Horizontal Translations
Build procedural fluency in reading $a$, $b$, $c$, $d$ from $y = a\sin(b(x-c)) + d$ — including the all-important "factor out $b$ first" move.
1. Quick recall
Three warm-ups. 1 mark each
Q1.1 Complete the general form: $y = $ ________ $\sin($ ________ $(x - $ ________ $)) + $ ________ .
Label what each parameter controls:
$a$ = ________________, $b$ = ________________, $c$ = ________________, $d$ = ________________.
Q1.2 Sign of $c$ tells you direction. Fill in:
If $c > 0$, the graph shifts to the ________.
If $c < 0$, the graph shifts to the ________.
Q1.3 In one sentence, explain why $y = \sin(x - c)$ shifts right by $c$ when $c > 0$ (counter-intuitive — the sign is "wrong way round").
2. Worked example — reading all four parameters
Problem. State the amplitude, period, phase shift, vertical shift, and range of $y = 2\sin\!\left(x - \dfrac{\pi}{3}\right) + 1$.
Step 1 — Compare with general form $y = a\sin(b(x-c)) + d$.
$a = 2$, $b = 1$, $c = \pi/3$, $d = 1$.
Reason: this expression is already in factored form — no factoring needed.
Step 2 — Compute amplitude and period.
Amplitude $= |a| = 2$.
Period $= 2\pi/|b| = 2\pi/1 = 2\pi$.
Reason: $|a|$, not $a$; $2\pi/|b|$, not $\pi/b$ (that's the tangent formula).
Step 3 — Read phase shift and vertical shift.
Phase shift: $c = \pi/3 > 0$, so shift $\pi/3$ to the right.
Vertical shift: $d = 1$, so up by $1$.
Step 4 — Find range.
Range $= [d - |a|, d + |a|] = [1 - 2, 1 + 2] = [-1, 3]$.
Conclusion. Amplitude $2$, period $2\pi$, phase shift $\pi/3$ right, vertical shift $1$ up, range $[-1, 3]$.
3. Faded example — fill in the missing steps
State the amplitude, period, phase shift, and vertical shift of $y = 3\cos\!\left(2x - \dfrac{\pi}{3}\right) - 4$. 5 marks
Step 1 — Factor out $b$ inside the bracket first.
$2x - \pi/3 = 2(x - $ ________ $)$.
Step 2 — Read off $a, b, c, d$.
$a = $ ________, $b = $ ________, $c = $ ________, $d = $ ________.
Step 3 — Amplitude and period.
Amplitude $= |a| = $ ________. Period $= 2\pi / |b| = $ ________.
Step 4 — Phase shift and vertical shift.
Phase shift: ________ to the ________. Vertical shift: ________ ________ (up/down).
Conclusion. Amplitude = ________, period = ________, phase shift = ________, vertical shift = ________.
4. Graduated practice
For each, state amplitude, period, phase shift (with direction), and vertical shift.
Foundation — already-factored form (4 questions)
| Q | Function | Amp / Period / Phase / Vert |
|---|---|---|
| 4.1 1 | $y = \sin(x - \pi/2)$ | |
| 4.2 1 | $y = \cos(x + \pi/4)$ | |
| 4.3 1 | $y = 2\sin x + 3$ | |
| 4.4 1 | $y = -\cos(x - \pi)$ |
Standard — factor out $b$ first (6 questions)
Show the factoring step.
4.5 $y = \sin(2x - \pi/3)$ — state all four parameters. 2 marks
4.6 $y = 3\cos(2x + \pi)$ — state all four parameters. 2 marks
4.7 $y = \sin(3x - \pi/2) + 2$ — state amplitude, period, phase shift, vertical shift, and range. 3 marks
4.8 $y = -2\sin(\pi x + \pi/2)$ — state all four parameters and the range. 3 marks
4.9 Write the equation of a sine curve with amplitude $4$, period $\pi$, phase shift $\pi/4$ right, and vertical shift $1$ down (in the form $y = a\sin(b(x-c)) + d$). 2 marks
4.10 Write the equation of a cosine curve with amplitude $0.5$, period $4\pi$, no phase shift, and vertical shift $2$ up. 2 marks
Extension — combine concepts (2 questions)
4.11 A sinusoidal curve has maximum value $5$, minimum value $-1$, and reaches its first positive maximum at $x = \pi/6$ (with period $2\pi$). Write the equation in the form $y = a\sin(b(x-c)) + d$ (using a phase shift of a sine, not a cosine). 3 marks
4.12 Show algebraically that $y = \cos x$ and $y = \sin(x + \pi/2)$ are the same function. Use this to convert $y = 3\cos(2(x - \pi/4)) - 1$ into pure sine form. 3 marks
5. Self-check the easy 3
Tick the first three once you've verified each one.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — General form
$y = a\sin(b(x - c)) + d$. $a$ = amplitude (and reflection if negative), $b$ controls period via $2\pi/|b|$, $c$ = phase shift (horizontal translation), $d$ = vertical shift (midline).
Q1.2 — Direction of phase shift
$c > 0$ shifts right. $c < 0$ shifts left. (Note: this refers to the factored form $\sin(b(x-c))$, not the unfactored $\sin(bx + k)$.)
Q1.3 — Why $\sin(x - c)$ shifts right by $c$
To get the same output as $\sin x$ at $x = 0$ from $\sin(x - c)$, we need $x - c = 0$, i.e. $x = c$. So the "$x = 0$ event" of the original graph now happens at $x = c$ — to the right if $c > 0$.
Q3 — Faded example: $y = 3\cos(2x - \pi/3) - 4$
Factor: $2x - \pi/3 = 2(x - \pi/6)$. $a = 3$, $b = 2$, $c = \pi/6$, $d = -4$. Amplitude $= 3$, period $= 2\pi/2 = \pi$. Phase shift = $\pi/6$ right, vertical shift = $4$ down. Amp $3$, period $\pi$, phase $\pi/6$ right, vert $4$ down.
Q4.1 — $y = \sin(x - \pi/2)$
Amplitude $1$, period $2\pi$, phase shift $\pi/2$ right, vertical shift $0$.
Q4.2 — $y = \cos(x + \pi/4)$
Rewrite $x + \pi/4 = x - (-\pi/4)$, so $c = -\pi/4$. Amplitude $1$, period $2\pi$, phase shift $\pi/4$ left, vertical shift $0$.
Q4.3 — $y = 2\sin x + 3$
Amplitude $2$, period $2\pi$, phase shift $0$, vertical shift $3$ up. Range $= [3 - 2, 3 + 2] = [1, 5]$.
Q4.4 — $y = -\cos(x - \pi)$
Amplitude $1$ (the minus reflects but $|a| = 1$), period $2\pi$, phase shift $\pi$ right, vertical shift $0$. Note: $-\cos(x - \pi) = \cos x$ (the two transformations cancel — extension observation).
Q4.5 — $y = \sin(2x - \pi/3)$
Factor: $2x - \pi/3 = 2(x - \pi/6)$. Amplitude $1$, period $\pi$, phase shift $\pi/6$ right, vertical shift $0$. (Trap 02: not $\pi/3$ right!)
Q4.6 — $y = 3\cos(2x + \pi)$
Factor: $2x + \pi = 2(x + \pi/2) = 2(x - (-\pi/2))$. Amplitude $3$, period $\pi$, phase shift $\pi/2$ left, vertical shift $0$.
Q4.7 — $y = \sin(3x - \pi/2) + 2$
Factor: $3x - \pi/2 = 3(x - \pi/6)$. Amplitude $1$, period $2\pi/3$, phase shift $\pi/6$ right, vertical shift $2$ up. Range $= [2 - 1, 2 + 1] = [1, 3]$.
Q4.8 — $y = -2\sin(\pi x + \pi/2)$
Factor: $\pi x + \pi/2 = \pi(x + 1/2) = \pi(x - (-1/2))$. Amplitude $|-2| = 2$ (with reflection in $x$-axis), period $2\pi/\pi = 2$, phase shift $1/2$ left, vertical shift $0$. Range $= [-2, 2]$.
Q4.9 — Build a sine equation
Amplitude $4 \Rightarrow a = 4$. Period $\pi \Rightarrow b = 2\pi/\pi = 2$. $c = \pi/4$. $d = -1$. $y = 4\sin(2(x - \pi/4)) - 1$.
Q4.10 — Build a cosine equation
$a = 0.5$, $b = 2\pi/(4\pi) = 1/2$, $c = 0$, $d = 2$. $y = 0.5\cos(x/2) + 2$.
Q4.11 — From max/min and max location
$a = (5 - (-1))/2 = 3$. $d = (5 + (-1))/2 = 2$. Period $2\pi \Rightarrow b = 1$. A sine reaches its first positive max at $x = c + \pi/2$ (i.e. quarter-period after the phase shift). Setting $c + \pi/2 = \pi/6 \Rightarrow c = \pi/6 - \pi/2 = -2\pi/6 = -\pi/3$. So $c = -\pi/3$ means shift $\pi/3$ left. $y = 3\sin(x + \pi/3) + 2$ (equivalently $3\sin(1 \cdot (x - (-\pi/3))) + 2$).
Q4.12 — Cosine as shifted sine
Identity. $\cos\theta = \sin(\pi/2 - \theta) = \sin(\theta + \pi/2)$ (the second equality uses $\sin$'s evenness about its peak; equivalently, $\cos\theta - \sin(\theta + \pi/2)$ can be expanded and simplifies to $0$). So $y = \cos\theta$ and $y = \sin(\theta + \pi/2)$ are the same function. Application. $y = 3\cos(2(x - \pi/4)) - 1 = 3\sin(2(x - \pi/4) + \pi/2) - 1$. Simplify the bracket: $2(x - \pi/4) + \pi/2 = 2x - \pi/2 + \pi/2 = 2x$. So $y = $ $3\sin(2x) - 1$. (Sanity check: cosine shifted right by $\pi/4$ with period $\pi$ — that's exactly an un-shifted sine with the same period.)