Math Core

Right-triangle definitions like "sine equals opposite over hypotenuse" break down once an angle is bigger than $90^\circ$ , because no right triangle has an angle that large. The unit circle fixes this. It defines sine and cosine for every angle, and it turns a table of values into a picture you can rebuild whenever you need it.

Sine and cosine as coordinates

The unit circle is the circle of radius $1$ centered at the origin, with equation $x^2 + y^2 = 1$ . Draw an angle $\theta$ in standard position. Its terminal side crosses the unit circle at exactly one point $P$ .

If $\theta$ is acute, drop a perpendicular from $P$ to the $x$ -axis. You get a right triangle with hypotenuse $1$ , so

\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{1} = x, \qquad \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{1} = y.

The cosine is the $x$ -coordinate of $P$ , and the sine is the $y$ -coordinate. That observation becomes the definition for all angles.

Definition

Sine and cosine of any angle

If the terminal side of $\theta$ meets the unit circle at the point $(x, y)$ , then

\cos\theta = x \qquad\text{and}\qquad \sin\theta = y.

Also, $\tan\theta = \dfrac{y}{x}$ whenever $x \ne 0$ .

Because every point on the unit circle has coordinates between $-1$ and $1$ , sine and cosine always lie in the interval $[-1, 1]$ . The quadrantal angles are easy to read off: at $\theta = 0$ the point is $(1, 0)$ , so $\cos 0 = 1$ and $\sin 0 = 0$ . At $\theta = \dfrac{\pi}{2}$ the point is $(0, 1)$ ; at $\theta = \pi$ it is $(-1, 0)$ ; at $\theta = \dfrac{3\pi}{2}$ it is $(0, -1)$ .

The special angles in Quadrant I

The angles $\dfrac{\pi}{6}$ , $\dfrac{\pi}{4}$ and $\dfrac{\pi}{3}$ come from the two special right triangles you met in Geometry.

A $45^\circ$ - $45^\circ$ - $90^\circ$ triangle with hypotenuse $1$ has legs $\dfrac{\sqrt{2}}{2}$ .
A $30^\circ$ - $60^\circ$ - $90^\circ$ triangle with hypotenuse $1$ has a short leg $\dfrac{1}{2}$ (opposite $30^\circ$ ) and a long leg $\dfrac{\sqrt{3}}{2}$ (opposite $60^\circ$ ).

Points on the unit circle for the angles π/6, π/4 and π/3.

$\theta$	$0$	$\dfrac{\pi}{6}$	$\dfrac{\pi}{4}$	$\dfrac{\pi}{3}$	$\dfrac{\pi}{2}$
$\cos\theta$	$1$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{1}{2}$	$0$
$\sin\theta$	$0$	$\dfrac{1}{2}$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{\sqrt{3}}{2}$	$1$

Tip

Notice the pattern in the sine row: $\dfrac{\sqrt{0}}{2}, \dfrac{\sqrt{1}}{2}, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{3}}{2}, \dfrac{\sqrt{4}}{2}$ . The cosine row is the same list backward. If you forget a value, rebuild it from this pattern, or ask yourself which coordinate is bigger: at $\dfrac{\pi}{3}$ the point is high and close to the $y$ -axis, so the sine is the larger value.

Reference angles and signs

Every other angle on the circle is a mirror image of a Quadrant I angle. The reference angle of $\theta$ is the acute angle between its terminal side and the $x$ -axis. A point at angle $\theta$ has the same coordinates as the point at its reference angle, except for the signs.

The signs depend only on the quadrant:

quadrant	I	II	III	IV
$\cos\theta$ ( $x$ )	$+$	$-$	$-$	$+$
$\sin\theta$ ( $y$ )	$+$	$+$	$-$	$-$

The angles π/6, 5π/6, 7π/6 and 11π/6 all have reference angle π/6. Their points are (±√3/2, ±1/2).

Evaluating sine and cosine of any angle

Find the quadrant of $\theta$ (use a coterminal angle if needed).
Find the reference angle.
Use the Quadrant I value for the reference angle, then attach the sign for that quadrant.

Worked example: Using reference angles

Find the exact value of each expression.

$\cos\dfrac{2\pi}{3}$
$\sin\dfrac{5\pi}{4}$
$\sin\left(-\dfrac{\pi}{6}\right)$

Solutions.

$\dfrac{2\pi}{3}$ is in Quadrant II, $\dfrac{\pi}{3}$ short of $\pi$ . The reference angle is $\dfrac{\pi}{3}$ , and cosine is negative in Quadrant II, so $\cos\dfrac{2\pi}{3} = -\dfrac{1}{2}$ .
$\dfrac{5\pi}{4}$ is $\dfrac{\pi}{4}$ past $\pi$ , in Quadrant III. Sine is negative there, so $\sin\dfrac{5\pi}{4} = -\dfrac{\sqrt{2}}{2}$ .
Rotating $\dfrac{\pi}{6}$ clockwise lands in Quadrant IV with reference angle $\dfrac{\pi}{6}$ . Sine is negative there, so $\sin\left(-\dfrac{\pi}{6}\right) = -\dfrac{1}{2}$ .

Common mistake

Don't mix up which coordinate is which. Cosine is $x$ (the horizontal position) and sine is $y$ (the height). Alphabetical order helps: $(x, y)$ matches $(\cos, \sin)$ , since c comes before s.

Circles of any radius

If a point $(x, y)$ lies on the terminal side of $\theta$ but not on the unit circle, scale it. Its distance from the origin is $r = \sqrt{x^2 + y^2}$ , and dividing by $r$ moves the point onto the unit circle. So

\cos\theta = \frac{x}{r}, \qquad \sin\theta = \frac{y}{r}, \qquad \tan\theta = \frac{y}{x}.

Worked example: A point on the terminal side

The terminal side of $\theta$ passes through $(-6, 8)$ . Find $\sin\theta$ and $\cos\theta$ .

First find the distance to the origin: $r = \sqrt{(-6)^2 + 8^2} = \sqrt{100} = 10$ . Then

\sin\theta = \frac{8}{10} = \frac{4}{5}, \qquad \cos\theta = \frac{-6}{10} = -\frac{3}{5}.

The signs make sense: the point is in Quadrant II, where $x$ is negative and $y$ is positive.

Practice

Practice 1

Find the exact value of $\sin\dfrac{\pi}{6}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Find the exact value of $\cos\dfrac{3\pi}{4}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Find the exact value of $\sin\left(-\dfrac{2\pi}{3}\right)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

What is the reference angle, in degrees, of $210^\circ$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

If $\sin\theta < 0$ and $\cos\theta > 0$ , in which quadrant does the terminal side of $\theta$ lie?

Practice 6

Give the coordinates of the point where the terminal side of $\dfrac{5\pi}{6}$ meets the unit circle.

Enter a point like (2, -3)

Practice 7

The terminal side of $\theta$ passes through $(-5, 12)$ . Find $\cos\theta$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

The point $\left(x, -\dfrac{3}{5}\right)$ is on the unit circle in Quadrant III. It lies on the terminal side of $\theta$ . Find $\cos\theta$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.