A Sine Shifted to the Right

Describe the graph of $y = \sin\left(x - \frac{\pi}{4}\right)$ compared with $y = \sin(x)$. Where does the first cycle start and where is its maximum?

Compare with $y = \sin(B(x - C))$: $x - \frac{\pi}{4}$ already has the form $1 \cdot (x - C)$ = $B = 1$, $C = \frac{\pi}{4}$

Read the direction: $C$ is positive = Shift $\frac{\pi}{4}$ to the right

Find the start of the cycle: The input is $0$ when $x - \frac{\pi}{4} = 0$ = Cycle starts at $x = \frac{\pi}{4}$

Move the maximum: $\sin(x)$ has its maximum at $\frac{\pi}{2}$, so add $\frac{\pi}{4}$: $\frac{\pi}{2} + \frac{\pi}{4}$ = Maximum at $x = \frac{3\pi}{4}$

Answer: The graph is $y = \sin(x)$ moved $\frac{\pi}{4}$ to the right. The first cycle starts at $x = \frac{\pi}{4}$ and its maximum $1$ is at $x = \frac{3\pi}{4}$.

A Plus Sign Means Left

Find the phase shift of $y = \cos\left(x + \frac{\pi}{3}\right)$ and the new position of the maximum that $\cos(x)$ has at $x = 0$.

Rewrite the plus as a minus: $x + \frac{\pi}{3} = x - \left(-\frac{\pi}{3}\right)$ = $C = -\frac{\pi}{3}$

Read the direction: $C$ is negative = Shift $\frac{\pi}{3}$ to the left

Move the maximum: $0 + \left(-\frac{\pi}{3}\right)$ = Maximum at $x = -\frac{\pi}{3}$

Answer: The phase shift is $-\frac{\pi}{3}$: the cosine moves $\frac{\pi}{3}$ to the left and its maximum is now at $x = -\frac{\pi}{3}$.

Factor Out B Before Reading the Shift

Find the phase shift and the period of $y = \sin\left(2x - \frac{\pi}{2}\right)$. Where does the first cycle start and end?

Factor out $B = 2$: $2x - \frac{\pi}{2} = 2\left(x - \frac{\pi}{4}\right)$ = $y = \sin\left(2\left(x - \frac{\pi}{4}\right)\right)$

Read the shift: $C = \frac{\pi}{4}$, positive = $\frac{\pi}{4}$ to the right (not $\frac{\pi}{2}$)

Find the period: $\frac{2\pi}{|B|} = \frac{2\pi}{2}$ = Period $= \pi$

Start and end of one cycle: Start at $C = \frac{\pi}{4}$, end at $\frac{\pi}{4} + \pi$ = From $x = \frac{\pi}{4}$ to $x = \frac{5\pi}{4}$

Answer: Phase shift $\frac{\pi}{4}$ to the right, period $\pi$. One cycle runs from $x = \frac{\pi}{4}$ to $x = \frac{5\pi}{4}$.

A Cosine With B and a Plus Sign

Find the phase shift and the period of $y = \cos\left(3x + \frac{\pi}{2}\right)$.

Factor out $B = 3$: $3x + \frac{\pi}{2} = 3\left(x + \frac{\pi}{6}\right)$, because $\frac{\pi}{2} \div 3 = \frac{\pi}{6}$ = $y = \cos\left(3\left(x + \frac{\pi}{6}\right)\right)$

Write it with a minus: $x + \frac{\pi}{6} = x - \left(-\frac{\pi}{6}\right)$ = $C = -\frac{\pi}{6}$

Read the direction: $C$ is negative = $\frac{\pi}{6}$ to the left

Find the period: $\frac{2\pi}{3}$ = Period $= \frac{2\pi}{3}$

Answer: The graph is $y = \cos(3x)$ moved $\frac{\pi}{6}$ to the left, with period $\frac{2\pi}{3}$.

Cosine as a Shifted Sine

Find a value of $C$ with $-\pi \le C \le \pi$ so that $\sin(x - C) = \cos(x)$ for every $x$.

Compare the maximums: $\cos(x)$ has its maximum at $x = 0$, $\sin(x)$ at $x = \frac{\pi}{2}$ = The sine peak must move from $\frac{\pi}{2}$ to $0$

Find the size and direction: From $\frac{\pi}{2}$ to $0$ is $\frac{\pi}{2}$ to the left = $C = -\frac{\pi}{2}$

Write the equation: $\sin\left(x - \left(-\frac{\pi}{2}\right)\right)$ = $\sin\left(x + \frac{\pi}{2}\right)$

Check one point: At $x = 0$: $\sin\left(\frac{\pi}{2}\right) = 1 = \cos(0)$ = It matches

Answer: $C = -\frac{\pi}{2}$, so $\cos(x) = \sin\left(x + \frac{\pi}{2}\right)$. Moving the sine $\frac{3\pi}{2}$ to the right also works, because that differs by one whole period $2\pi$.

Mistake: Reading $y = \sin\left(x - \frac{\pi}{3}\right)$ as a shift to the left

Why: The minus sign looks like it should point left, as on a number line. But the curve reaches each value later, when $x - \frac{\pi}{3}$ equals what $x$ used to be.

Correct: Minus inside the bracket means right: $\sin(x - C)$ moves $C$ to the right, $\sin(x + C)$ moves $C$ to the left.

Mistake: Saying the shift of $y = \sin\left(2x - \frac{\pi}{2}\right)$ is $\frac{\pi}{2}$

Why: The number $\frac{\pi}{2}$ is subtracted from $2x$, not from $x$. The shift belongs to $x$ alone.

Correct: Factor out $B$: $\sin\left(2\left(x - \frac{\pi}{4}\right)\right)$. The shift is $\frac{c}{B} = \frac{\pi}{4}$ to the right.

Mistake: Thinking a phase shift changes the period or the height

Why: Moving a graph sideways does not stretch it. Only $B$ changes the period and only $A$ changes the amplitude.

Correct: After a phase shift the curve has the same shape, period and amplitude. Only the place where each cycle starts changes.

Mistake: Thinking $y = \sin(x - 2\pi)$ is a different graph from $y = \sin(x)$

Why: Sine repeats every $2\pi$, so moving it by a whole period puts every wave exactly on the next one.

Correct: Shifts that differ by a whole period give the same graph: $\sin(x - 2\pi) = \sin(x)$, and $\sin\left(x + \frac{\pi}{2}\right) = \sin\left(x - \frac{3\pi}{2}\right)$.

Predicting High Tide

Harbor masters model the water level with a shifted sine. The phase shift tells them when the water passes the mean level, and from there they find high tide.

If the water level is $h(t) = 1.5\sin\left(\frac{\pi}{6}(t - 2)\right)$ meters above the mean level, with $t$ in hours after midnight, the water rises through the mean level at $t = 2$, at 2:00.

Alternating Current

Power grids in Europe run at 50 Hz. Two voltages with the same frequency can be out of step, and engineers measure that difference as a phase shift in seconds.

The voltage $V_1 = \sin(100\pi t)$ has period $\frac{2\pi}{100\pi} = \frac{1}{50}$ second. A second voltage $V_2 = \sin\left(100\pi t - \frac{\pi}{2}\right)$ reaches each peak a little later.

Noise-Cancelling Headphones

A microphone picks up the noise, and the headphones play the same wave shifted by half a period. Where the noise goes up, the copy goes down, and the two cancel.

For the noise $y = \sin(x)$ the copy is $y = \sin(x - \pi)$, which equals $-\sin(x)$. Their sum is $0$ everywhere.

Q: Why is it called a phase shift?

A: The phase is the position inside one cycle (start, peak, zero, trough). A phase shift moves every point of the cycle by the same amount, so the whole wave starts earlier or later.

Q: Is the phase shift of $\sin(2x - \pi)$ equal to $\pi$ or $\frac{\pi}{2}$?

A: It is $\frac{\pi}{2}$ to the right. Factor out $2$: $\sin(2x - \pi) = \sin\left(2\left(x - \frac{\pi}{2}\right)\right)$. Some books call $\pi$ the phase angle, but the distance the graph moves on the $x$-axis is $\frac{\pi}{2}$.

Q: Can a graph have more than one correct phase shift?

A: Yes. Adding or subtracting a whole period gives the same graph, so $\sin\left(x - \frac{\pi}{3}\right)$ and $\sin\left(x + \frac{5\pi}{3}\right)$ are identical. Usually you give the shift with the smallest size.

Q: Does the order of stretching and shifting matter?

A: Yes. $\sin(2(x - 1))$ and $\sin(2x - 1)$ are different: the first is shifted $1$ to the right, the second only $\frac{1}{2}$. Writing the function as $\sin(B(x - C))$ avoids the confusion.