Phase Shift of a Sine Wave

A phase shift moves a wave horizontally without changing its shape, amplitude, or period. The function $y = \sin(x - \varphi)$ is the same wave as $y = \sin x$, just shifted to the right by $\varphi$.

The Rule

Inside the sine function, subtracting $\varphi$ shifts the graph to the right by $\varphi$. Adding $\varphi$ shifts it to the left. (This is opposite of what your intuition might expect — "minus moves right" is one of the trickiest things to internalize in graph transformations.)

Step-by-Step Solution

Compare $y_1 = \sin x$ and $y_2 = \sin(x - \pi/4)$.

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Step 1 — Find the zero crossings of $y_1 = \sin x$.

$\sin x = 0 \;\Longrightarrow\; x = 0,\, \pi,\, 2\pi,\, 3\pi,\, \ldots$

Step 2 — Find the zero crossings of $y_2 = \sin(x - \pi/4)$.

Set the argument equal to multiples of $\pi$:

$x - \dfrac{\pi}{4} = 0,\, \pi,\, 2\pi,\, \ldots$

$x = \dfrac{\pi}{4},\, \pi + \dfrac{\pi}{4},\, 2\pi + \dfrac{\pi}{4},\, \ldots$

Each zero of $y_2$ is shifted right by $\pi/4 \approx 0.785$ compared to the corresponding zero of $y_1$.

Step 3 — Find the maxima of both.

$y_1$ peaks where $\sin x = 1$, at $x = \pi/2,\, 5\pi/2,\, \ldots$

$y_2$ peaks where $\sin(x - \pi/4) = 1$, i.e. $x - \pi/4 = \pi/2$, so $x = \pi/2 + \pi/4 = 3\pi/4$.

So the peak shifts from $x = \pi/2 \approx 1.571$ to $x = 3\pi/4 \approx 2.356$ — a shift of exactly $\pi/4 \approx 0.785$ to the right. ✓

Step 4 — Tabulate values at a few common $x$ values.

• At $x = 0$: $y_1 = 0$, $y_2 = \sin(-\pi/4) = -\sqrt{2}/2 \approx -0.707$
• At $x = \pi/4$: $y_1 = \sqrt{2}/2 \approx 0.707$, $y_2 = \sin(0) = 0$ (zero!)
• At $x = \pi/2$: $y_1 = 1$ (peak), $y_2 = \sin(\pi/4) = \sqrt{2}/2$
• At $x = 3\pi/4$: $y_1 = \sqrt{2}/2$, $y_2 = \sin(\pi/2) = 1$ (peak!)

You can clearly see $y_2$ "lagging behind" $y_1$ by exactly $\pi/4$ at every step.

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Answer: The graph $y = \sin(x - \pi/4)$ is identical to $y = \sin x$ but shifted to the right by $\pi/4$ (about 0.785 units). Every zero, every peak, and every trough is delayed by exactly that amount.

In general, $y = \sin(x - \varphi)$ shifts the wave right by $\varphi$, and $y = \sin(x + \varphi)$ shifts it left by $\varphi$.

Try It

• Slide the phase shift $\varphi$ widget — watch the pink wave slide horizontally relative to the cyan reference.
• A horizontal arrow shows the magnitude and direction of the shift.
• At $\varphi = \pi/2 \approx 1.571$, the shifted wave becomes $\sin(x - \pi/2) = -\cos x$.
• At $\varphi = \pi$, the shifted wave is $\sin(x - \pi) = -\sin x$ — flipped upside down.