Math Core

A handful of angles appear again and again in trigonometry: multiples of $30^\circ$ and $45^\circ$ . Their sines and cosines have exact values built from $\dfrac{1}{2}$ , $\dfrac{\sqrt{2}}{2}$ and $\dfrac{\sqrt{3}}{2}$ . In this lesson you'll build the whole unit circle from two special triangles, then use reference angles to evaluate any of these angles in seconds, without a calculator.

The first quadrant

Remember that the point where the terminal side of $\theta$ meets the unit circle is $(\cos\theta, \sin\theta)$ . In Quadrant I, dropping a perpendicular from that point to the $x$ -axis makes a right triangle with hypotenuse 1. So the two special right triangles give the coordinates directly.

The 30-60-90 triangle. Its sides are in the ratio $1 : \sqrt{3} : 2$ . Scale it so the hypotenuse is $1$ : the short leg is $\dfrac{1}{2}$ and the long leg is $\dfrac{\sqrt{3}}{2}$ . The short leg is opposite the $30^\circ$ angle.

A 30° angle cuts off a 30-60-90 triangle with hypotenuse 1.

At $30^\circ$ the triangle is wide and short, so the point is $\left(\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}\right)$ .
At $60^\circ$ the triangle is tall and narrow, so the legs swap: $\left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)$ .

The 45-45-90 triangle. Its legs are equal and its sides are in the ratio $1 : 1 : \sqrt{2}$ . With hypotenuse $1$ , each leg is $\dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$ , so the point at $45^\circ$ is $\left(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}\right)$ .

$\theta$	$0$	$\frac{\pi}{6}$ ( $30^\circ$ )	$\frac{\pi}{4}$ ( $45^\circ$ )	$\frac{\pi}{3}$ ( $60^\circ$ )	$\frac{\pi}{2}$ ( $90^\circ$ )
$\cos\theta$	$1$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	$0$
$\sin\theta$	$0$	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	$1$

Tip

A memory pattern: the sines from $0^\circ$ to $90^\circ$ are $\dfrac{\sqrt{0}}{2}, \dfrac{\sqrt{1}}{2}, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{3}}{2}, \dfrac{\sqrt{4}}{2}$ . The cosines are the same list in reverse. And a quick sanity check: as the angle grows in Quadrant I, the point rises (sine gets bigger) and moves left (cosine gets smaller).

Reflecting into the other quadrants

The unit circle is symmetric across both axes. That means every special angle in another quadrant has a point that is a mirror image of a Quadrant I point: the same numbers, possibly with different signs.

The Quadrant I angle that matches is the reference angle: the acute angle between the terminal side and the $x$ -axis. For example, $150^\circ$ is $30^\circ$ short of $180^\circ$ , so its reference angle is $30^\circ$ . Its point is the reflection of the $30^\circ$ point across the $y$ -axis.

150° and 30° have mirror-image points. Only the sign of x changes.

Reference angles in radians work the same way. For $\theta$ between $0$ and $2\pi$ :

terminal side in	reference angle
Quadrant I	$\theta$
Quadrant II	$\pi - \theta$
Quadrant III	$\theta - \pi$
Quadrant IV	$2\pi - \theta$

Doing this for every multiple of $\dfrac{\pi}{6}$ and $\dfrac{\pi}{4}$ fills in the full unit circle: $16$ special angles in all.

The 16 special angles on the unit circle, in radians.

Look for families. Every angle with denominator $6$ ( $\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$ ) has a $30^\circ$ reference angle. Every angle with denominator $4$ has a $45^\circ$ reference angle. Every angle with denominator $3$ has a $60^\circ$ reference angle.

Evaluating a trig function at a special angle

If needed, add or subtract $360^\circ$ (or $2\pi$ ) to get a coterminal angle between $0$ and $360^\circ$ .
Find the reference angle and the Quadrant I value that goes with it.
Attach the sign that fits the quadrant: $x$ (cosine) is negative on the left, $y$ (sine) is negative below.
For the other functions, use $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ and the reciprocals.

Worked example: Quadrants II and III

Find the exact values of $\cos 150^\circ$ and $\sin\dfrac{5\pi}{4}$ .

$\cos 150^\circ$ . The reference angle is $180^\circ - 150^\circ = 30^\circ$ , and $\cos 30^\circ = \dfrac{\sqrt{3}}{2}$ . The point is in Quadrant II, where $x$ is negative, so $\cos 150^\circ = -\dfrac{\sqrt{3}}{2}$ .

$\sin\dfrac{5\pi}{4}$ . The reference angle is $\dfrac{5\pi}{4} - \pi = \dfrac{\pi}{4}$ , and $\sin\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}$ . The point is in Quadrant III, where $y$ is negative, so $\sin\dfrac{5\pi}{4} = -\dfrac{\sqrt{2}}{2}$ .

Worked example: Tangent and secant

Find the exact values of $\tan\dfrac{5\pi}{3}$ and $\sec\dfrac{2\pi}{3}$ .

$\tan\dfrac{5\pi}{3}$ . The reference angle is $2\pi - \dfrac{5\pi}{3} = \dfrac{\pi}{3}$ , in Quadrant IV. So the point is $\left(\dfrac{1}{2}, -\dfrac{\sqrt{3}}{2}\right)$ , and

\tan\frac{5\pi}{3} = \frac{y}{x} = \frac{-\sqrt{3}/2}{1/2} = -\sqrt{3}.

$\sec\dfrac{2\pi}{3}$ . The reference angle is $\pi - \dfrac{2\pi}{3} = \dfrac{\pi}{3}$ , in Quadrant II. So $\cos\dfrac{2\pi}{3} = -\dfrac{1}{2}$ , and $\sec\dfrac{2\pi}{3} = \dfrac{1}{-1/2} = -2$ .

Worked example: Large and negative angles

Find the exact values of $\cos\dfrac{11\pi}{3}$ and $\sin(-225^\circ)$ .

$\cos\dfrac{11\pi}{3}$ . Subtract $2\pi = \dfrac{6\pi}{3}$ : $\dfrac{11\pi}{3} - \dfrac{6\pi}{3} = \dfrac{5\pi}{3}$ . That's in Quadrant IV with reference angle $\dfrac{\pi}{3}$ , and cosine is positive there: $\cos\dfrac{11\pi}{3} = \dfrac{1}{2}$ .

$\sin(-225^\circ)$ . Add $360^\circ$ : $-225^\circ + 360^\circ = 135^\circ$ . That's in Quadrant II with reference angle $45^\circ$ , and sine is positive there: $\sin(-225^\circ) = \dfrac{\sqrt{2}}{2}$ .

Common mistake

Don't mix up which coordinate is which. Sine is the $y$ -coordinate (up and down), and cosine is the $x$ -coordinate (left and right). A common slip is writing $\sin 60^\circ = \dfrac{1}{2}$ . Picture the point at $60^\circ$ : it's high up, so its $y$ -coordinate is the bigger one, $\dfrac{\sqrt{3}}{2}$ .

Tangent values

Since $\tan\theta = \dfrac{y}{x}$ , the Quadrant I tangents are

\tan 30^\circ = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}, \qquad \tan 45^\circ = 1, \qquad \tan 60^\circ = \sqrt{3}.

Tangent is the slope of the terminal side. A shallow line through the origin has a small slope, and a steep line has a big one. That's an easy way to remember that $\tan 30^\circ$ is the small one and $\tan 60^\circ$ the large one.