A wave that starts at its peak is different from a wave that starts at zero — even if they have the same amplitude and period. This difference is called a phase shift. In this lesson, you will learn how horizontal translations affect trigonometric graphs, how to identify phase shifts from equations, and how phase shifts appear in real-world phenomena like sound, light, and tides.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Consider the equations $y = \sin x$ and $y = \sin(x - \frac{\pi}{2})$. If you substitute $x = \frac{\pi}{2}$ into the second equation, what value do you get? How does this compare to $y = \sin x$ when $x = 0$? What do you think this means about the graph of $y = \sin(x - \frac{\pi}{2})$?
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
Wrong: The sine and cosine functions have a period of 360° for all transformations.
Right: The period is affected by horizontal dilation; y = sin(nx) has period 360°/n, not 360°.
📚 Core Content
In the general form $y = a\sin\big(b(x - c)\big) + d$, the parameter $c$ controls the horizontal translation, also called the phase shift.
It may seem backwards that $x - c$ shifts right. But think about what value of $x$ gives the same output as the unshifted graph at $x = 0$. For $y = \sin(x - c)$, we need $x - c = 0$, so $x = c$. The "starting behaviour" that used to happen at $x = 0$ now happens at $x = c$ — which is to the right if $c$ is positive.
For $y = a\sin\big(b(x - c)\big) + d$ or $y = a\cos\big(b(x - c)\big) + d$:
| Parameter | Effect |
|---|---|
| $a$ | Amplitude = $|a|$. Reflects in the $x$-axis if $a < 0$. |
| $b$ | Period = $\frac{2\pi}{|b|}$. Stretches or compresses horizontally. |
| $c$ | Phase shift = $c$. Translates horizontally. |
| $d$ | Vertical shift = $d$. Translates vertically; midline becomes $y = d$. |
The range is always $[d - |a|, d + |a|]$, regardless of the phase shift or period.
🧮 Worked Examples
🧪 Activities
1 $y = 3\sin\left(x - \frac{\pi}{4}\right)$
Type your answer:
Answer in your workbook.
2 $y = 2\cos(2x + \pi) - 1$
Type your answer:
Answer in your workbook.
3 $y = \sin\left(\frac{x}{2} - \frac{\pi}{6}\right) + 2$
Type your answer:
Answer in your workbook.
1 Amplitude 2, period $2\pi$, phase shift $\frac{\pi}{3}$ right, vertical shift 1 up
Type your answer:
Answer in your workbook.
2 Amplitude 4, period $\pi$, phase shift $\frac{\pi}{4}$ left, vertical shift 2 down
Type your answer:
Answer in your workbook.
3 Amplitude 1, period $4\pi$, no phase shift, vertical shift 3 up
Type your answer:
Answer in your workbook.
Earlier you were asked about $y = \sin(x - \frac{\pi}{2})$.
When $x = \frac{\pi}{2}$: $y = \sin\left(\frac{\pi}{2} - \frac{\pi}{2}\right) = \sin 0 = 0$. This is the same value as $y = \sin x$ at $x = 0$. So the graph of $y = \sin(x - \frac{\pi}{2})$ is the sine graph shifted $\frac{\pi}{2}$ units to the right. In fact, this graph is identical to $y = -\cos x$.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
8. For $y = 2\cos\left(2x - \frac{\pi}{3}\right) + 1$, find the amplitude, period, phase shift, and range. Show all working. 4 MARKS
Type your answer below:
Answer in your workbook.
9. Sketch $y = \sin\left(x + \frac{\pi}{6}\right)$ for $-\frac{\pi}{6} \leq x \leq \frac{11\pi}{6}$. Label the phase shift and the coordinates of all maximum and minimum points. 4 MARKS
Describe your sketch below:
Sketch in your workbook.
10. Two sound waves are modelled by $y = \sin x$ and $y = \sin(x - \frac{\pi}{2})$. Explain what the phase difference between these two waves means in a real-world context. 3 MARKS
Type your answer below:
Answer in your workbook.
1. Amplitude = 3, Period = $2\pi$, Phase shift = $\frac{\pi}{4}$ right, Vertical shift = 0.
2. Rewrite: $y = 2\cos\left(2\left(x + \frac{\pi}{2}\right)\right) - 1$. Amplitude = 2, Period = $\pi$, Phase shift = $\frac{\pi}{2}$ left, Vertical shift = 1 down.
3. Rewrite: $y = \sin\left(\frac{1}{2}\left(x - \frac{\pi}{3}\right)\right) + 2$. Amplitude = 1, Period = $4\pi$, Phase shift = $\frac{\pi}{3}$ right, Vertical shift = 2 up.
1. $y = 2\sin\left(x - \frac{\pi}{3}\right) + 1$
2. $y = 4\sin\left(2\left(x + \frac{\pi}{4}\right)\right) - 2$
3. $y = \sin\left(\frac{x}{2}\right) + 3$
1. A — $y = \sin(x - \pi)$ shifts right by $\pi$.
2. A — Period of $\cos(2x)$ is $\pi$.
3. A — Factor: $\sin(2x - \frac{\pi}{3}) = \sin\left(2(x - \frac{\pi}{6})\right)$. Phase shift = $\frac{\pi}{6}$ right.
4. A — Amplitude = 3, vertical shift = $-1$, so range = $[-4, 2]$.
5. A — $y = \sin(x + \frac{\pi}{2})$ is the same as $y = \cos x$.
Q8 (4 marks): Rewrite: $y = 2\cos\left(2\left(x - \frac{\pi}{6}\right)\right) + 1$ [1]. Amplitude = 2 [0.5], Period = $\pi$ [0.5], Phase shift = $\frac{\pi}{6}$ right [1]. Range = $[-1, 3]$ [1].
Q9 (4 marks): Phase shift = $\frac{\pi}{6}$ left [1]. Max at $x = \frac{\pi}{3}$ (value 1) [1]. Min at $x = \frac{4\pi}{3}$ (value $-1$) [1]. Correct sine wave shape over the domain [1].
Q10 (3 marks): The second wave is shifted $\frac{\pi}{2}$ to the right [1]. This represents a phase difference of $\frac{\pi}{2}$ radians (a quarter of a cycle) [1]. In sound, this means the second wave reaches its peak a quarter cycle later than the first, changing how they interfere [1].
Sprint through questions on phase shifts and horizontal translations of trig graphs. Pool: lessons 1–12.
Tick when you've finished all activities and checked your answers.