Mathematics Advanced • Year 11 • Module 2 • Lesson 12
Phase Shifts and Horizontal Translations
Practise HSC-style writing on factored form, building equations from graph features, and the algebraic equivalence of sine and cosine via phase shift.
1. Short-answer questions
1.1 For $y = 2\sin\!\left(x - \dfrac{\pi}{3}\right) + 1$, state the amplitude, period, phase shift (with direction), vertical shift, and range. 3 marks Band 3
1.2 Consider $y = \sin(2x - \pi/3)$.
(a) Rewrite in the form $y = \sin(b(x - c))$ by factoring out $b$.
(b) State the period and phase shift (with direction).
(c) State the coordinates of the first positive $x$-intercept (i.e. smallest $x > 0$ where $y = 0$). 4 marks Band 3-4
1.3 A sinusoidal curve has amplitude $3$, period $\pi$, is shifted $\pi/6$ to the right, and has a vertical shift of $2$ units down. Write its equation in the form $y = a\sin(b(x - c)) + d$, then verify by computing the $y$-value at $x = \pi/6$. 4 marks Band 4
Stuck on 1.3? At the phase-shift point $x = c$, a pure sine takes the value $d$ (the midline) and is rising.2. Extended response
2.1 Consider the function $f(x) = \sin(2x + \pi/2)$ and its proposed alternative form $g(x) = \cos(2x)$.
(a) Rewrite $f(x)$ in the form $a\sin(b(x - c)) + d$ by factoring $2$ out of the bracket, and state the amplitude, period, phase shift (direction), and vertical shift.
(b) Use the identity $\sin(\theta + \pi/2) = \cos\theta$ to show that $f(x) = g(x)$ for all $x$. Then explain in your own words why the graph of $f$ and the graph of $g$ are the same curve despite being written with different "starting" trig functions.
(c) A student writes "$y = \sin(2x - \pi/3)$ has phase shift $\pi/3$ right". Evaluate this claim, give the correct phase shift, and explain explicitly which of the lesson's Traps the student has fallen into. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 3 marks
• 1 mark — factors $2x + \pi/2 = 2(x + \pi/4)$ correctly.
• 1 mark — states amplitude $1$, period $\pi$, vertical shift $0$.
• 1 mark — states phase shift $\pi/4$ left (must give direction).
Part (b) — 3 marks
• 1 mark — invokes the identity $\sin(\theta + \pi/2) = \cos\theta$.
• 1 mark — substitutes $\theta = 2x$ to get $\sin(2x + \pi/2) = \cos(2x) = g(x)$.
• 1 mark — explains that sine and cosine are themselves phase-shifted versions of each other, so any sinusoid can be written in either form.
Part (c) — 2 marks
• 1 mark — identifies the error: $2x - \pi/3 = 2(x - \pi/6)$, so the phase shift is $\pi/6$, not $\pi/3$.
• 1 mark — names Trap 02 (not factoring out $b$ before reading $c$).
Your response:
Stuck on (b)? Sine shifted left by $\pi/2$ becomes cosine — this is one of the most-used identities in HSC.How did this worksheet feel?
What I'll revisit before next class:
1.1 — $y = 2\sin(x - \pi/3) + 1$ (3 marks)
Sample response. Amplitude $2$, period $2\pi$, phase shift $\pi/3$ right, vertical shift $1$ up, range $[-1, 3]$.
Marking notes. 1 mark for amplitude + period, 1 mark for phase shift with direction, 1 mark for vertical shift + range. Common error: stating phase shift as $\pi/3$ without "right" — costs 0.5.
1.2 — $y = \sin(2x - \pi/3)$ (4 marks)
Sample response.
(a) $2x - \pi/3 = 2(x - \pi/6)$, so $y = \sin(2(x - \pi/6))$ [1].
(b) Period $= 2\pi/2 = \pi$, phase shift $= \pi/6$ right [1].
(c) The factored form is a sine of $2(x - \pi/6)$. First $x$-intercept of unshifted $\sin(2x)$ in $x > 0$ is at $x = 0$ (but that's not strictly positive), then $x = \pi/2$. With phase shift $\pi/6$ right, intercepts at $x = \pi/6 + n\pi/2$. The smallest positive is at $n = 0$: $x = \pi/6$ [2].
Marking notes. (a) 1 mark for the factoring. (b) 1 mark — both period and phase shift correct. (c) 2 marks — 1 for setting up the intercept pattern, 1 for identifying $\pi/6$ as the first positive. Common error: confusing "first intercept" with "first positive intercept" — students often pick $x = 0$ or a negative value.
1.3 — Build equation and verify (4 marks)
Sample response. $a = 3$. Period $\pi \Rightarrow b = 2\pi/\pi = 2$. $c = \pi/6$. $d = -2$. So $y = $ $3\sin(2(x - \pi/6)) - 2$ [3].
Verification at $x = \pi/6$: $y = 3\sin(2(\pi/6 - \pi/6)) - 2 = 3\sin(0) - 2 = 0 - 2 = -2$, which equals $d$ (the midline). ✓ — this is the expected value at the phase-shift point (where the sine wave is on its midline going up) [1].
Marking notes. 3 marks for building the equation (1 each for $a$, $b$, $c+d$). 1 mark for the verification — must show $y = -2$ at $x = \pi/6$ and note that this is the midline value (the conceptual point of the verification).
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a) — Factoring $f$. The expression $2x + \pi/2$ factors as $2(x + \pi/4) = 2(x - (-\pi/4))$. So
$f(x) = 1 \cdot \sin(2(x - (-\pi/4))) + 0$. [1 mark — factoring]
Therefore amplitude $= 1$, period $= 2\pi/2 = \pi$, vertical shift $= 0$ [1 mark], and phase shift $= -\pi/4$, i.e. $\pi/4$ to the left (positive $c$ in the unfactored expression became negative once factored). [1 mark — direction]
Part (b) — Equivalence with $g(x) = \cos(2x)$. The identity $\sin(\theta + \pi/2) = \cos\theta$ holds for all $\theta$ [1 mark]. Substituting $\theta = 2x$:
$f(x) = \sin(2x + \pi/2) = \cos(2x) = g(x)$. [1 mark — substitution]
The two functions $f$ and $g$ are identical because sine and cosine are themselves phase-shifted versions of each other (cosine = sine shifted $\pi/2$ left), so any sinusoidal curve can be written equivalently as a phase-shifted sine or a phase-shifted cosine. The graph itself doesn't change — only the algebraic description does. [1 mark — conceptual]
Part (c) — The student's error. The student is incorrect. In $\sin(2x - \pi/3)$, the coefficient of $x$ inside the bracket is $2$, not $1$. To read the phase shift, the bracket must first be factored:
$2x - \pi/3 = 2(x - \pi/6)$. So the phase shift is $\pi/6$ right, not $\pi/3$ right. [1 mark — correct answer]
This is exactly Trap 02 in the lesson: "Not factoring out $b$ before identifying $c$". The constant subtracted inside the bracket has units of angle ($\pi/3$ rad), but the phase shift is in units of $x$; the conversion factor is $1/b$. [1 mark — Trap 02 named]
Total: 8/8.
Band descriptors for marker.
Band 3: Reads off parameters without factoring; states an identity in (b) but doesn't apply it; identifies the student's answer is wrong but gives no correct value. ≈ 3-4 marks.
Band 4: Factors correctly in (a); cites the identity and substitutes in (b); gives the correct $\pi/6$ in (c) but doesn't name Trap 02. ≈ 5-6 marks.
Band 5: All parts correct including direction in (a), conceptual explanation in (b), and explicit Trap 02 reference in (c). ≈ 7 marks.
Band 6: All of the above plus the dimensional argument in (c) (that the "constant inside bracket" has units of angle, and dividing by $b$ converts to units of $x$). 8/8.