C. Trigonometric Functions · Sine Functions

Horizontal Translations: f(x) = sin(x − d)

C2.6 C2.7About 65 minutes

Learning goals

I can describe the effect of d on the graph of f(x) = sin(x − d).
I can tell which way the graph moves from the sign inside the brackets.
I can sketch f(x) = sin(x − d) by moving the five key points and state its domain and range.
I can find d from the location of a maximum, minimum or zero.

Words to know

horizontal translation
A shift of every point of a graph left or right by the same amount.

phase shift
Another name for the horizontal translation of a periodic function.

d (in sin(x − d))
The horizontal translation: right if d > 0, left if d < 0.

Starting the wave later

Two friends get on the same Ferris wheel, one minute apart. Their heights follow exactly the same pattern, but the second friend's graph is the first friend's graph delayed: slid to the right. A sideways slide of a graph is a horizontal translation. For sine functions it is often called a phase shift.

Remember from Grade 10

The graph of y = (x − h)² is y = x² moved h units right. Notice the surprise: a minus sign inside the brackets moves the graph to the right, and y = (x + 3)² moves it 3 units left. Sine works the same way.

The role of d in f(x) = sin(x − d)

To get the same output as sin x had at angle x, the new function needs an input that is d larger: sin((x + d) − d) = sin x. So every point moves horizontally by d:

(x, y) → (x + d, y)

If d > 0, as in y = sin(x − 60°), the graph moves d to the right.
If d < 0, the graph moves left. For example, y = sin(x + 45°) = sin(x − (−45°)), so d = −45° and the graph moves 45° to the left.

Read the sign carefully: x − 60° means right 60°; x + 45° means left 45°.

What changes and what stays the same?

The x-values of the zeros, maximums and minimums all move by d. One cycle now starts at x = d (instead of 0°): key points (d, 0), (90° + d, 1), (180° + d, 0), (270° + d, −1), (360° + d, 0).
The amplitude (1), period (360°), axis (y = 0), domain and range (−1 ≤ y ≤ 1) do not change.

y = sin(x − 60°) (solid) is y = sin x (dashed) moved 60° to the right. Each arrow shows a key point sliding 60° right; the maximum moves from 90° to 150°.

Worked example

Example 1: A shift to the right

Sketch y = sin(x − 60°). Give its key points for one cycle, its maximum and minimum points, its zeros from 0° to 360°, and its range.

d = 60°, so the graph moves 60° right: (x, y) → (x + 60°, y).
Key points: (0° + 60°, 0) = (60°, 0), (150°, 1), (240°, 0), (330°, −1), (420°, 0).
Maximum point (150°, 1); minimum point (330°, −1).
Zeros from 0° to 360°: 60° and 240°. (The next zero, 420°, is beyond 360°; the one before, 60° − 180° = −120°, is below 0°.)
Range: unchanged, {y ∈ ℝ | −1 ≤ y ≤ 1}. Check one point: at x = 150°, sin(150° − 60°) = sin 90° = 1 ✓.

Key points (60°, 0), (150°, 1), (240°, 0), (330°, −1), (420°, 0); zeros 60° and 240°; range −1 ≤ y ≤ 1.

Worked example

Example 2: A shift to the left

Describe the transformation in y = sin(x + 45°) and find its key points and its y-intercept.

x + 45° = x − (−45°), so d = −45°: the graph moves 45° left. (x, y) → (x − 45°, y).
Key points: (−45°, 0), (45°, 1), (135°, 0), (225°, −1), (315°, 0).
y-intercept: y = sin(0° + 45°) = sin 45° ≈ 0.707.

Left 45°; key points (−45°, 0), (45°, 1), (135°, 0), (225°, −1), (315°, 0); y-intercept about 0.707.

y = sin(x + 45°) (solid) is y = sin x (dashed) moved 45° to the left. The maximum is now at 45° and a cycle starts at −45°.

Worked example

Example 3: Finding d from a graph

A curve has the shape of y = sin x (amplitude 1, axis y = 0), but its first maximum to the right of the y-axis is at (120°, 1). Write an equation in the form y = sin(x − d).

For y = sin x the maximum is at 90°. Here it is at 120°.
The graph moved 120° − 90° = 30° right, so d = 30°.
Check: at x = 120°, sin(120° − 30°) = sin 90° = 1 ✓.

y = sin(x − 30°). (Because the period is 360°, y = sin(x + 330°) gives the same graph, but the smallest shift is usually used.)

Two useful facts

Shifting by a full period changes nothing: sin(x − 360°) = sin x.
Shifting by half a period flips the wave: sin(x − 180°) = −sin x. (Check at 90°: sin(−90°) = −1 = −sin 90°.) So the same graph can sometimes be described in more than one way.

Investigate with technology

Graph Y1 = sin(X) and Y2 = sin(X − D) for D = 30, 90, −60 and 180 (degree mode, window −90 ≤ x ≤ 450). For each, record where the first maximum is. You should find it at 90° + D.

Just for fun

y = sin(x − 90°) arrived at class 90° late.

"Sorry," it said. "My phase shifted."

Hands-on activity

Two people, one wheel

You need: two coins (or bottle caps), a dinner plate, a piece of paper, a pencil and a timer.

Trace the plate on paper and mark 0°, 90°, 180° and 270° around it (0° at the right, going counterclockwise).
Place coin A at 0° and coin B at 270° (at the bottom of the plate). Move both coins counterclockwise together, 30° at a time, and record each coin's height above the centre line in a table (above = +, below = −; use the plate radius as 1).
Plot both sets of points on one set of axes (angle turned on the x-axis).
Describe how B's graph is related to A's. Write B's equation as y = sin(x − d).

What to look for: coin B starts at the bottom, so its graph is A's graph moved 90° to the right: y = sin(x − 90°). (Moving left 270° gives the same graph, y = sin(x + 270°), because the period is 360°.)

Your turn

Describe the transformation that maps y = sin x onto y = sin(x − 20°).

Describe the transformation that maps y = sin x onto y = sin(x + 90°).

Give the key points of one cycle of y = sin(x − 30°).

State the amplitude, period and range of y = sin(x + 70°).

Find the y-intercept of y = sin(x − 30°).

A sine curve (amplitude 1, axis y = 0) has its first maximum to the right of the y-axis at 135°. Write its equation as y = sin(x − d).

More practice: warm-up

In y = sin(x − 50°), what is d?

In y = sin(x + 50°), what is d?

Which way does y = sin(x + 10°) move?

Where is the maximum of y = sin(x − 40°) between 0° and 360°?

Write the mapping rule for y = sin(x − 25°).

More practice: on your own

Find the zeros of y = sin(x − 75°) from 0° to 360°.

Find the maximum and minimum points of y = sin(x + 120°) from 0° to 360°.

Evaluate y = sin(x − 45°) at x = 0°, 45°, 135° and 225°.

Show that y = sin(x − 180°) and y = −sin x give the same value at x = 30°, 90° and 200°.

A student says y = sin(x + 30°) is shifted 30° right because of the plus sign. Correct the student.

Challenge

Find two different equations of the form y = sin(x − d), one with 0° < d < 360° and one with −360° < d < 0°, for a sine curve whose maximum is at (200°, 1).

Solve sin(x − 30°) = 0.5 for 0° ≤ x ≤ 360°.

Explain why a horizontal translation does not change the range of a sine function.

Check your understanding

Quiz

How is y = sin(x − 40°) related to y = sin x?

How is y = sin(x + 15°) related to y = sin x?

Where is the maximum of y = sin(x − 20°) for 0° ≤ x ≤ 360°?

Which property changes in y = sin(x − d)?

Which equation has a minimum at (300°, −1)?

Which equation gives the same graph as y = sin x?

For parents and teachers

Big idea: The role of d in f(x) = sin(x − d), a horizontal translation or phase shift (C2.6), and sketching the result with its domain and range (C2.7).

How to help: Have your teen say the shift out loud before sketching: "x minus 60 means 60 to the right." Then check one point by substituting: at the new maximum x-value, the expression inside the brackets should equal 90°.

Common difficulties:

The sign. sin(x − 30°) moves right, sin(x + 30°) moves left.
Shifting the graph up or down instead of sideways.
Forgetting that zeros, maximums and minimums all move by the same amount.