Unit 6, lesson 46

The Sine Function

How sine turns every real radian input into the height of a point on the unit circle, and how that height produces the sine wave with its range, period, odd symmetry, zeros and amplitude.

  • Theory part 91
  • Practice part 92
  • The Six Trigonometric Functions

A point on the rim of a turning wheel rises and falls in a steady rhythm. Track only its height as the wheel turns and you get a wave that repeats forever. That height, for a wheel of radius one, is the sine of the angle turned, and the wave is the graph of the sine function. Calculus treats sine as a function of a real number like any other, so this page builds it that way.

Key terms

Sine as a unit-circle height

Definition For a real number xx, let P(x)P(x) be the point where the terminal side of the angle of xx radians in standard position meets the unit circle. The sine of xx is the vertical coordinate of that point,

P(x)=(cos⁡x,sin⁡x).P(x) = (\cos x, \sin x).

This defines a function sin⁡\sin whose domain is all real numbers.

In plain words Turn through the angle xx and look at where you land on the circle. The sine is how high that point sits above the horizontal axis, and it is negative when the point is below the axis.

Example The angle 7π6\frac{7\pi}{6} lands in quadrant III with reference angle π6\frac{\pi}{6}. The point there is below the axis at distance 12\frac{1}{2}, so

sin⁡7π6=−12.\sin \frac{7\pi}{6} = -\frac{1}{2}.

Radian input

Definition The input of sin⁡\sin is a real number read as an angle in radians. On the unit circle the input xx is the signed length of the arc from (1,0)(1, 0) to P(x)P(x), counterclockwise for x>0x > 0 and clockwise for x<0x < 0. If an angle is given in degrees, it must first be converted using 180∘=π180^\circ = \pi radians. The calculus rules for sine, such as the derivative of sin⁡x\sin x being cos⁡x\cos x, hold only with radian input.

In plain words The number you feed to sine is a distance walked around the unit circle. Degrees are a different unit, and the same number in degrees names a different angle.

Example An angle of 30∘30^\circ is π6\frac{\pi}{6} radians, so its sine is 12\frac{1}{2}. The real number 3030, read as radians, is almost five full turns, and

sin⁡30≈−0.988.\sin 30 \approx -0.988.

The two answers are nowhere near each other, which is why a calculator in degree mode gives wrong answers in calculus. See radians and angle measurement for the conversion.

Range

Definition The range of sin⁡\sin is the closed interval [−1,1][-1, 1]. Every real xx satisfies −1≤sin⁡x≤1-1 \le \sin x \le 1, and every number yy in [−1,1][-1, 1] equals sin⁡x\sin x for some real xx. The largest value 11 is reached exactly at x=π2+2πkx = \frac{\pi}{2} + 2\pi k and the smallest value −1-1 exactly at x=3π2+2πkx = \frac{3\pi}{2} + 2\pi k, for integers kk.

In plain words A point on a circle of radius one can never be more than one unit above or below the center, and it passes through every height in between.

Example The equation sin⁡x=2\sin x = 2 has no real solution, because 22 is outside [−1,1][-1, 1]. The equation sin⁡x=−1\sin x = -1 has the solutions x=3π2+2πkx = \frac{3\pi}{2} + 2\pi k, and the only one in [0,2π)[0, 2\pi) is 3π2\frac{3\pi}{2}.

Period

Definition A function ff is periodic with period p>0p > 0 if x+px + p is in the domain whenever xx is, and f(x+p)=f(x)f(x + p) = f(x) for every xx in the domain. For sine, adding 2πk2\pi k to the input adds kk full turns, so for every real xx and every integer kk

sin⁡(x+2πk)=sin⁡x.\sin(x + 2\pi k) = \sin x.

The least positive period of sine is 2π2\pi. To see that nothing smaller works, suppose p>0p > 0 is a period. Then

sin⁡(π2+p)=sin⁡π2=1.\sin\left(\frac{\pi}{2} + p\right) = \sin \frac{\pi}{2} = 1.

By the range facts above, sine equals 11 only at the inputs π2+2πk\frac{\pi}{2} + 2\pi k, so p=2πkp = 2\pi k for a positive integer kk, and therefore p≥2πp \ge 2\pi.

In plain words After one full turn the point is back where it started, so the heights repeat. One full turn is the shortest shift that repeats everything.

Example The input 25π6\frac{25\pi}{6} is 4π+π64\pi + \frac{\pi}{6}, two full turns past π6\frac{\pi}{6}. So

sin⁡25π6=sin⁡π6=12.\sin \frac{25\pi}{6} = \sin \frac{\pi}{6} = \frac{1}{2}.

Odd symmetry

Definition A function ff is odd if −x-x is in its domain whenever xx is and f(−x)=−f(x)f(-x) = -f(x) for all such xx. Sine is odd,

sin⁡(−x)=−sin⁡xfor every real x.\begin{gathered} \sin(-x) = -\sin x \\[0.4em] \text{for every real } x. \end{gathered}

The reason is that P(−x)P(-x) is the reflection of P(x)P(x) across the horizontal axis, which keeps the horizontal coordinate and flips the sign of the vertical one. The graph of an odd function is symmetric about the origin.

In plain words Turning the same amount clockwise instead of counterclockwise lands you at the mirror point below the axis, so the height has the same size and the opposite sign.

Example The value sin⁡π3=32\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} gives at once

sin⁡(−π3)=−32.\sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}.

Zeros

Definition The zeros of sine are the real numbers xx with sin⁡x=0\sin x = 0. They are exactly

x=kπ,k an integer.x = k\pi, \qquad k \text{ an integer}.

These are the inputs whose point is (1,0)(1, 0) or (−1,0)(-1, 0), the two places where the unit circle crosses the horizontal axis.

In plain words The height is zero only when the point is at the far right or the far left of the circle, which happens every half turn.

Example The zeros of sine in [−2π,2π][-2\pi, 2\pi] are −2π-2\pi, −π-\pi, 00, π\pi and 2π2\pi, five in all. The input π2\frac{\pi}{2} is not a zero, since there the point is at the top and sin⁡π2=1\sin \frac{\pi}{2} = 1.

Sine graph

Definition The graph of sine is the set of points (x,sin⁡x)(x, \sin x) for all real xx. One full cycle runs over [0,2π][0, 2\pi], and the rest of the graph is copies of that cycle shifted by multiples of 2π2\pi. The five landmarks of the cycle are the zeros, the peak and the trough,

xx00π2\frac{\pi}{2}π\pi3π2\frac{3\pi}{2}2π2\pi
sin⁡x\sin x001100−1-100

Sine increases on [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] and decreases on [π2,3π2]\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]. A finite table of values only illustrates the graph. The smooth shape between the points comes from the point moving continuously around the circle, not from the table.

In plain words Unroll the circle along the horizontal axis and plot the height of the moving point above each input. The landmarks fix the skeleton, and the motion of the point fills in the curve.

Example Adding the angles from the special triangles gives more points on the first half of the cycle.

xx00π6\frac{\pi}{6}π4\frac{\pi}{4}π3\frac{\pi}{3}π2\frac{\pi}{2}2π3\frac{2\pi}{3}5π6\frac{5\pi}{6}π\pi
sin⁡x\sin x0012\frac{1}{2}22\frac{\sqrt{2}}{2}32\frac{\sqrt{3}}{2}1132\frac{\sqrt{3}}{2}12\frac{1}{2}00

The values climb quickly near 00 and slowly near the peak, then fall back in the mirror pattern, which is why the wave is rounded at the top and steep where it crosses the axis.

Amplitude

Definition For real numbers A≠0A \ne 0 and B≠0B \ne 0, the function y=Asin⁡(Bx)y = A \sin(Bx) has amplitude ∣A∣\lvert A \rvert, range [−∣A∣,∣A∣][-\lvert A \rvert, \lvert A \rvert] and least positive period

2π∣B∣.\frac{2\pi}{\lvert B \rvert}.

The amplitude is the distance from the midline y=0y = 0 to a peak. Each landmark (x,y)(x, y) of sin⁡x\sin x moves to (xB,Ay)\left(\frac{x}{B}, Ay\right).

In plain words The factor AA stretches the wave vertically, and a negative AA also flips it. The factor BB squeezes the wave horizontally, so a full cycle takes less input when ∣B∣\lvert B \rvert is bigger.

Example For y=3sin⁡(2x)y = 3\sin(2x) the amplitude is 33 and the period is 2π2=π\frac{2\pi}{2} = \pi. Halve each landmark input and triple each output.

xx00π4\frac{\pi}{4}π2\frac{\pi}{2}3π4\frac{3\pi}{4}π\pi
3sin⁡(2x)3\sin(2x)003300−3-300

The wave rises to 33, returns to 00, drops to −3-3 and is back at 00 after one period of length π\pi.

Check yourself

  1. Find sin⁡11π6\sin \frac{11\pi}{6} exactly.
Show the answer

The angle is in quadrant IV with reference angle 2π−11π6=π62\pi - \frac{11\pi}{6} = \frac{\pi}{6}. The point is below the axis, so the sine is negative, and sin⁡11π6=−12\sin \frac{11\pi}{6} = -\frac{1}{2}.

  1. Give the amplitude and least positive period of y=−4sin⁡(πx3)y = -4\sin\left(\frac{\pi x}{3}\right), and list its landmarks over one period starting at x=0x = 0.
Show the answer

The amplitude is ∣−4∣=4\lvert -4 \rvert = 4 and the period is

2ππ/3=6.\frac{2\pi}{\pi / 3} = 6.

Each landmark input of sine is multiplied by 3π\frac{3}{\pi} and each output by −4-4, which gives (0,0)(0, 0), (1.5,−4)(1.5, -4), (3,0)(3, 0), (4.5,4)(4.5, 4) and (6,0)(6, 0). The negative factor makes the wave dip first.

  1. Find every xx in [0,4π][0, 4\pi] with sin⁡x=1\sin x = 1.
Show the answer

Sine equals 11 exactly at x=π2+2πkx = \frac{\pi}{2} + 2\pi k. The values in [0,4π][0, 4\pi] come from k=0k = 0 and k=1k = 1, so x=π2x = \frac{\pi}{2} and x=5π2x = \frac{5\pi}{2}.

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