Phase Shifts and Horizontal Translations
A wave that starts at its peak is different from a wave that starts at zero — even if they share 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 they appear in sound, light, and tides.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
What is the amplitude and period of $y = \sin x$? What happens when you replace $x$ with $x - \frac{\pi}{2}$? Predict how the graph changes before you read on.
Every transformed sine or cosine function can be written in general form. The four parameters $a$, $b$, $c$, and $d$ each control a different transformation. Learn to read them quickly and you can sketch any sinusoidal graph in seconds.
The general form $y = a\sin(b(x - c)) + d$ contains four parameters. $|a|$ is the amplitude, $\frac{2\pi}{|b|}$ is the period, $c$ is the phase shift (horizontal translation), and $d$ is the vertical shift. The range is always $[d - |a|,\ d + |a|]$.
Key insight: A positive $c$ shifts the graph right; a negative $c$ shifts it left. Always factor out $b$ before reading $c$.
Key facts
- The general form of transformed sine and cosine functions
- How to identify amplitude, period, phase shift, and vertical shift
- The relationship between phase shift and horizontal translation
Concepts
- Why $y = \sin(x - c)$ shifts right by $c$ units
- How phase shifts affect the starting point of a wave
- Why phase shifts are important in real-world periodic phenomena
Skills
- Sketch transformed sine and cosine graphs with phase shifts
- Find amplitude, period, phase shift, and vertical shift from an equation
- Write the equation of a sinusoidal graph given its features
In the general form $y = a\sin(b(x - c)) + d$, the parameter $c$ controls the horizontal translation, also called the phase shift.
Three transformations: amplitude (blue, taller), period (orange, compressed), phase shift (green, shifted right by $\frac{\pi}{4}$)
- If $c > 0$, the graph shifts right by $c$ units.
- If $c < 0$, the graph shifts left by $|c|$ units.
Why right instead of left?
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.
Phase shift = $c$ in $y = a\sin(b(x-c))+d$; positive $c$ shifts right, negative shifts left; Always factor out $b$ before reading the phase shift: $y = \sin(2x - \frac{\pi}{3}) = \sin(2(x - \frac{\pi}{6}))$, so phase shift = $\frac{\pi}{6}$ right
Pause — copy the phase shift rule (positive $c$ = shift right; negative $c$ = shift left) and the must-factor-out-$b$ warning with the example $\sin(2x - \frac{\pi}{3}) = \sin(2(x - \frac{\pi}{6}))$ into your book.
Did you get this? True or false: in $y = \sin(x + \frac{\pi}{3})$, the graph is shifted $\frac{\pi}{3}$ units to the right.
We just saw that phase shift $c$ moves the graph left or right, but only after factoring $b$ out of the brackets to read it correctly. That raises a question: how do all four parameters — amplitude, period, phase shift, vertical shift — work together in one formula? This card answers it → the complete general form $y = a\sin(b(x-c))+d$ and the range formula $[d-|a|, d+|a|]$.
For $y = a\sin(b(x - c)) + d$ or $y = a\cos(b(x - c)) + d$:
| 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 $y = d$. |
Finding the range
The range is always $[d - |a|,\ d + |a|]$, regardless of the phase shift or period.
General form summary: $y = a\sin(b(x-c))+d$ — four parameters, four transformations; Range: $[d-|a|,\ d+|a|]$ — phase shift and period do not affect the range
Pause — copy the four-parameter summary $y = a\sin(b(x-c))+d$ and the range shortcut $[d-|a|, d+|a|]$ (phase shift and period don't affect range) into your book.
Quick check: What is the phase shift of $y = \sin\left(2x - \frac{\pi}{3}\right)$?
Worked examples · 3 in a row
For $y = 2\sin\left(x - \frac{\pi}{3}\right) + 1$, state the amplitude, period, phase shift, and vertical shift.
Sketch $y = \cos\left(x + \frac{\pi}{4}\right)$ for $-\frac{\pi}{4} \leq x \leq \frac{7\pi}{4}$ and label the phase shift.
We just saw how to read all four parameters from $y = a\sin(b(x-c))+d$ and how phase shift and period don't affect range. That raises a question: can we run the process in reverse — given the features (amplitude, period, phase shift), can we reconstruct the equation? This card answers it → two worked examples showing both directions of the feature–equation conversion.
A sinusoidal graph has amplitude 3, period $\pi$, is shifted $\frac{\pi}{6}$ to the right, and has a vertical shift of 2 units down. Write its equation in the form $y = a\sin(b(x - c)) + d$.
WE1: $y = 2\sin(x - \frac{\pi}{3}) + 1$ — amplitude 2, period $2\pi$, phase shift $\frac{\pi}{3}$ right, vertical shift 1 up; WE2: $y = \cos(x + \frac{\pi}{4})$ — rewrite as $\cos(x - (-\frac{\pi}{4}))$, shift $\frac{\pi}{4}$ left
Pause — copy the two worked results: $y = 2\sin(x - \frac{\pi}{3}) + 1$ (reading features from equation) and the rewrite $\cos(x + \frac{\pi}{4}) = \cos(x - (-\frac{\pi}{4}))$ — shift is $\frac{\pi}{4}$ left into your book.
Fill in the blank: A sinusoidal function with amplitude 5, period $4\pi$, and no phase shift or vertical shift is $y = 5\sin\!\left(\underline{\hspace{40px}}\,x\right)$.
We just saw how to read off or reconstruct all four parameters, including the counterintuitive direction of phase shift. That raises a question: what are the specific errors students most often make with phase shifts in the exam? This card answers it → Trap 1: $\sin(x+c)$ shifts LEFT (opposite to the sign); Trap 2: never read phase shift before factoring out $b$.
Trap 1: $\sin(x + c)$ shifts LEFT; $\sin(x - c)$ shifts RIGHT — the direction is opposite to the sign; Trap 2: Always factor out $b$ first: $\sin(2x - \frac{\pi}{2}) = \sin(2(x - \frac{\pi}{4}))$, phase shift $= \frac{\pi}{4}$
Pause — copy both traps: Trap 1 ($\sin(x+c)$ shifts LEFT — sign is opposite), Trap 2 (factor out $b$ first: $\sin(2x - \frac{\pi}{2}) = \sin(2(x - \frac{\pi}{4}))$, phase shift $= \frac{\pi}{4}$ right) into your book.
Odd one out: Which equation has a phase shift to the RIGHT?
Quick-fire practice · 5 reps
$y = 3\sin\left(x - \frac{\pi}{4}\right)$ — state amplitude, period, phase shift, and vertical shift.
$y = 2\cos(2x + \pi) - 1$ — state all four features (factor out 2 first!).
$y = \sin\left(\frac{x}{2} - \frac{\pi}{6}\right) + 2$ — state all four features.
Sketch $y = \sin(x + \frac{\pi}{6})$ for $-\frac{\pi}{6} \leq x \leq \frac{11\pi}{6}$ — describe key features.
Write the equation: amplitude 2, period $2\pi$, phase shift $\frac{\pi}{3}$ right, vertical shift 1 up.
Teach it back: Explain to a classmate why you must factor out $b$ before reading the phase shift from an equation like $y = \sin(3x - \pi)$.
Earlier you were asked about $y = \sin(x - \frac{\pi}{2})$.
When $x = \frac{\pi}{2}$: $y = \sin(\frac{\pi}{2} - \frac{\pi}{2}) = \sin 0 = 0$. This matches $y = \sin x$ at $x = 0$. So the graph is the sine wave shifted $\frac{\pi}{2}$ units to the right. In fact, this graph is identical to $y = -\cos x$.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. For $y = 2\cos\left(2x - \frac{\pi}{3}\right) + 1$, find the amplitude, period, phase shift, and range. Show all working. (4 marks)
View comprehensive answer
Rewrite: $y = 2\cos\left(2\left(x - \frac{\pi}{6}\right)\right) + 1$ [1 mark].
Amplitude = $|2| = 2$ [0.5]. Period = $\frac{2\pi}{2} = \pi$ [0.5]. Phase shift = $\frac{\pi}{6}$ right [1].
Vertical shift = 1, so range = $[1 - 2,\ 1 + 2] = [-1, 3]$ [1].
Q2. 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)
View comprehensive answer
Phase shift = $\frac{\pi}{6}$ left [1].
Maximum at $x = \frac{\pi}{3}$ (value 1) [1]. Minimum at $x = \frac{4\pi}{3}$ (value $-1$) [1].
Correct sine wave shape over the domain, starting at $(-\frac{\pi}{6}, 0)$ and ending at $(\frac{11\pi}{6}, 0)$ [1].
Q3. 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)
View comprehensive answer
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].
Comprehensive answers (click to reveal)
Drill 1: Amplitude = 3, Period = $2\pi$, Phase shift = $\frac{\pi}{4}$ right, Vertical shift = 0.
Drill 2: $y = 2\cos(2(x + \frac{\pi}{2})) - 1$. Amplitude = 2, Period = $\pi$, Phase shift = $\frac{\pi}{2}$ left, Vertical shift = 1 down.
Drill 3: $y = \sin(\frac{1}{2}(x - \frac{\pi}{3})) + 2$. Amplitude = 1, Period = $4\pi$, Phase shift = $\frac{\pi}{3}$ right, Vertical shift = 2 up.
Drill 4: Phase shift = $\frac{\pi}{6}$ left. Max at $(\frac{\pi}{3}, 1)$, min at $(\frac{4\pi}{3}, -1)$. Standard sine shape shifted left.
Drill 5: $y = 2\sin(x - \frac{\pi}{3}) + 1$.
SAQ1 (4 marks): Rewrite [1], amplitude = 2 [0.5], period = $\pi$ [0.5], phase shift = $\frac{\pi}{6}$ right [1], range = $[-1, 3]$ [1].
SAQ2 (4 marks): Phase shift = $\frac{\pi}{6}$ left [1], max at $\frac{\pi}{3}$ [1], min at $\frac{4\pi}{3}$ [1], correct shape [1].
SAQ3 (3 marks): Shift $\frac{\pi}{2}$ right [1], quarter cycle difference [1], real-world interference explanation [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering phase shift and trigonometric graph questions.
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