Two waves where there was one
The graph of y = sin x makes one whole wave between 0° and 360°: up to 1 at 90°, back to 0 at 180°, down to −1 at 270° and back to 0 at 360°. The length of one whole wave is the period, so the period of y = sin x is 360°.
Now draw y = sin 2x on the same axes. It makes two whole waves between 0° and 360°. Each wave is the same height as before, from −1 up to 1, but it is squeezed into half the width. So the period of y = sin 2x is 180°.
y = sin 2x in gold and y = sin x behind it, with one square across for every 90°. The gold wave peaks at 45° and finishes its first wave at 180°, where y = sin x is only halfway through its own.
Why the 2 halves the period
The sine is taken of the angle inside the bracket. In y = sin 2x that angle is 2x, and it grows twice as fast as x does. When x = 45°, the angle is 2 × 45° = 90°, so y = sin 90° = 1, the first peak. When x = 90°, the angle is 180°; when x = 135°, it is 270°; and when x = 180°, it is 360°.
The sine finishes one whole wave when the angle inside it has grown by 360°. Here 2x grows by 360° while x grows by only 180°. So one wave of y = sin 2x takes 180° of x, and between 0° and 360° there is room for two.
This is a horizontal stretch. Multiplying x by 2 inside the function squeezes the graph toward the y-axis by a factor of ½: every point moves to half its distance from the y-axis, and the heights do not change.
The period is 360° ÷ b
The same argument works for any number b multiplying x. The angle bx grows by 360° when x grows by 360° ÷ b, so the period of y = sin bx is 360° ÷ b. The same is true of y = cos bx.
For y = sin 3x the period is 360° ÷ 3 = 120°, so three whole waves fit between 0° and 360°. For y = cos 4x it is 360° ÷ 4 = 90°, and four waves fit. A number less than 1 stretches the wave instead: for y = sin ½x the period is 360° ÷ ½ = 720°, so between 0° and 360° there is only half a wave.
A bigger b means more waves in the same space, so a shorter period. The number in front does not affect the period at all: y = 3 sin 2x runs between −3 and 3, and its period is still 360° ÷ 2 = 180°.
The rule also runs backwards. A sine wave with a period of 60° has 360° ÷ b = 60°, so , and its equation is y = sin 6x.
Sliding the wave along
Now subtract a number inside the bracket: y = sin (x − 90°). This does not change the shape or the period. It slides the whole wave along the x-axis.
The plain wave y = sin x starts on the axis and rises, at x = 0°. The new wave does the same where the angle inside the bracket is 0°, which is where x − 90° = 0, so at x = 90°. Each point of y = sin x arrives 90° later. So y = sin (x − 90°) is the graph of y = sin x moved 90° to the right.
The minus sign moves the wave to the right, not to the left, because x has to be 90° bigger before the angle inside the bracket reaches each value it had before. A plus sign moves it to the left: y = sin (x + 90°) is y = sin x moved 90° to the left, and that is the graph of cosine.
The distance a wave is slid along is called the phase shift. For y = sin (x − 90°) the phase shift is 90° to the right.
y = sin (x − 90°) in gold and y = sin x behind it, with one square across for every 90°. The gold wave starts rising at 90° and peaks at 180°, each 90° later than y = sin x.
the peak is where the input x − c is 0, so it sits at x = c: the graph moved RIGHT by 2
Set c = 3 and find where the peak lands
The same rule on a simpler graph, a single bump with its peak at 0. In y = f(x − c) the peak is where x − c = 0, which is x = c. Drag c and the bump moves the same way as c: right for a positive c, even though the bracket shows a minus sign.
Four numbers, four jobs
Put everything together and a sine wave can be written y = a sin (b(x − c)) + d. Each number does one job. a is the amplitude, how far the wave rises above its principal axis. 360° ÷ b is the period. c is the phase shift, how far the wave is slid to the right. d is the level of the principal axis, y = d.
Take y = 2 sin (3(x − 20°)) + 1. The amplitude is 2 and the principal axis is y = 1, so the wave runs from 1 − 2 = −1 up to 1 + 2 = 3. The period is 360° ÷ 3 = 120°. The phase shift is 20° to the right.
The first crest after the shift comes where the angle inside the sine is 90°: 3(x − 20°) = 90°, so x − 20° = 30° and x = 50°. Check: y = 2 sin (3 × 30°) + 1 = 2 sin 90° + 1 = 2 + 1 = 3, the greatest value. The next crest is one period later, at 50° + 120° = 170°.
Factor the bracket first
An equation is not always written with b outside the bracket. In y = sin (2x − 60°), the phase shift is not 60°. Take out the factor of 2 first: 2x − 60° = 2(x − 30°), so y = sin (2(x − 30°)).
Now read it: the period is 360° ÷ 2 = 180° and the phase shift is 30° to the right. Check where the wave starts rising: the angle 2x − 60° is 0° when 2x = 60°, so at x = 30°.
The usual mistakes
Multiplying instead of dividing. The period of y = sin 3x is 360° ÷ 3 = 120°, not 360° × 3 = 1080°. A bigger b squeezes more waves in, so the period gets shorter.
Sliding the wave the wrong way. y = sin (x − 90°) moves 90° to the right. The minus sign inside the bracket does not mean left.
Reading the shift from an unfactored bracket. In y = sin (2x − 60°) the shift is 30°, because 2x − 60° = 2(x − 30°).
Counting crests and troughs as separate waves. One whole wave holds one crest and one trough, so y = sin 3x draws three waves between 0° and 360°, not six.
Letting a or d change the period. They change the height and the level of the wave. Only b changes the period, and only c slides it along.
A model of daylight
The application below models the hours of daylight D in a city, x months after June 21, by D = a cos (bx)° + c. In this model c is added outside the cosine, so it is the level of the principal axis, the d of y = a sin (b(x − c)) + d. There is no phase shift: the model starts at June 21, the longest day, and the graph of cosine starts at its peak.
The model works out as D = 4.5 cos (30x)° + 12, which has a period of 360 ÷ 30 = 12 months, one year. Part (b) then needs the times when D = 14.25, which come down to cos (30x)° = ½. The exact value cos 60° = ½ gives one answer, 30x = 60, so x = 2.
The second answer comes from the symmetry of the cosine graph. It is symmetrical about each of its peaks, so the curve crosses the same level as far before a peak as it does after it. The next peak is at x = 12, one period on, so the curve also crosses D = 14.25 at x = 12 − 2 = 10.
D = 4.5 cos (30x)° + 12, with x in months after June 21. It peaks at 16.5 hours at x = 0 and x = 12, and falls to 7.5 hours at x = 6. The dashed line D = 14.25 crosses the curve at x = 2 and x = 10, two months either side of the peak at x = 12.
Worked example: Hours of Daylight in a City in Northern Europe Through the Year
Question In a city in northern Europe the longest day, June 21, has 16.5 hours of daylight and the shortest, December 21, has 7.5 hours. The number of hours of daylight, D, x months after June 21 is modeled by D = a cos (bx)° + c, taking each month as one twelfth of a year. (a) Find a, b and c. (b) Between which dates does the city have more than 14.25 hours of daylight?
1.The principal axis is halfway between the greatest and least values: c = 16.5 + 7.52 = 12 hours.
The principal axis is halfway between the longest and shortest days: c = 16.5 + 7.52 = 12. 2.The amplitude is how far the curve rises above the principal axis: a = 16.5 − 7.52 = 4.5 hours. At x = 0, June 21, the daylight is at its greatest, and the graph of cosine starts at its peak, so the model is a cosine with a positive.
The amplitude is half the difference: a = 16.5 − 7.52 = 4.5 hours. The curve is at a peak at x = 0, so the model is a cosine. 3.One cycle is a year of 12 months, and bx must grow by 360 in that time, so b = 36012 = 30. (a) a = 4.5, b = 30 and c = 12, so D = 4.5 cos (30x)° + 12. Check: at x = 6, December 21, D = 4.5 cos 180° + 12 = 12 − 4.5 = 7.5 hours.
(a) A year of 12 months is one cycle, so b = 36012 = 30, and D = 4.5 cos (30x)° + 12. 4.Set D = 14.25: 4.5 cos (30x)° = 2.25, so cos (30x)° = 12. The principal value is 30x = 60, so x = 2, which is August 21.
cos (30x)° = 12 first where 30x = 60, so x = 2: August 21. 5.The graph is symmetrical about each of its peaks. The next peak is at x = 12, June 21 of the next year, so the curve also crosses D = 14.25 two months before it, at x = 10, which is April 21, and it is above the line from x = 10 to x = 14. (b) The city has more than 14.25 hours of daylight from April 21 to August 21, four months of the year. Check: at x = 11, May 21, D = 4.5 cos 330° + 12 = 15.9 hours.
(b) The graph is symmetrical about its peak at x = 12, so it also crosses at x = 10, April 21: above the line from April 21 to August 21.
Answer: (a) a = 4.5, b = 30, c = 12; (b) from April 21 to August 21, 4 months, for x from 10 to 14
Common mistakes
- Using a sine: D = 4.5 sin (30x)° + 12 gives 12 hours at x = 0, but June 21 is the longest day, where the graph must be at its peak. The graph of sine starts on its principal axis; the graph of cosine starts at its peak.
- Giving the dates from August 21 to April 21. Between those dates the curve is below the line D = 14.25; the stretch above it lies on either side of the peak at June 21.