Sine and Cosine Function Values — AP® Precalculus Notes

The exam expects exact values of sine and cosine at the angles you meet most often, with no calculator. This page shows how to get each one in a few seconds from two triangles and the signs of the quadrants, and how to use them on circles of any size.

The first quadrant: three triangles of hypotenuse 1

Put the angle $\frac{\pi}{6}$ ( $30^\circ$ ) on the unit circle and drop a segment from the point to the $x$ -axis. You get a 30-60-90 triangle with hypotenuse 1. Its sides are $1 : \sqrt{3} : 2$ divided by 2: the height (across from $30^\circ$ ) is $\frac{1}{2}$ and the base is $\frac{\sqrt{3}}{2}$ . So the point is $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ .

At $\frac{\pi}{3}$ ( $60^\circ$ ) the same triangle stands on its other leg, and the coordinates swap: $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ . At $\frac{\pi}{4}$ the 45-45-90 triangle with hypotenuse 1 has equal legs $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ .

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

A quick check: the ray at $\frac{\pi}{6}$ is close to the $x$ -axis, so the point is far to the right and not very high. The bigger value, $\frac{\sqrt{3}}{2} \approx 0.87$ , is the $x$ -coordinate.

Every other quadrant: reference angles and signs

These are the special angles: every multiple of $\frac{\pi}{6}$ and $\frac{\pi}{4}$ . Each one makes the same triangle with the $x$ -axis as one of the three first-quadrant angles. The acute angle between the terminal ray and the $x$ -axis is the reference angle. The point has the same coordinates as at the reference angle, up to sign, and the quadrant gives the signs.

A circle of radius 1 centered at the origin with a dot at every multiple of pi/6 and pi/4 from 0 to 2 pi, each labeled with its angle in radians. The first-quadrant points are labeled with their coordinates: at pi/6, (square root of 3 over 2, 1/2); at pi/4, (square root of 2 over 2, square root of 2 over 2); at pi/3, (1/2, square root of 3 over 2). A table to the right lists cosine and sine at 0, pi/6, pi/4, pi/3 and pi/2: cosine 1, root 3 over 2, root 2 over 2, 1/2, 0 and sine 0, 1/2, root 2 over 2, root 3 over 2, 1. The other quadrants repeat these values with signs set by the quadrant. — **Figure 1.** All sixteen special angles and the first-quadrant values they repeat. LevlPrep original diagram.

Quadrant	Angles between	Reference angle of $\theta$	Signs of $(\cos\theta, \sin\theta)$
I	$0$ and $\frac{\pi}{2}$	$\theta$	$(+, +)$
II	$\frac{\pi}{2}$ and $\pi$	$\pi - \theta$	$(-, +)$
III	$\pi$ and $\frac{3\pi}{2}$	$\theta - \pi$	$(-, -)$
IV	$\frac{3\pi}{2}$ and $2\pi$	$2\pi - \theta$	$(+, -)$

Worked example 1. Find $\cos\frac{4\pi}{3}$ and $\sin\frac{4\pi}{3}$ .

$\frac{4\pi}{3}$ is between $\pi$ and $\frac{3\pi}{2}$ : Quadrant III. Its reference angle is $\frac{4\pi}{3} - \pi = \frac{\pi}{3}$ , where the point is $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ .

In Quadrant III both coordinates are negative: $\cos\frac{4\pi}{3} = -\frac{1}{2}$ and $\sin\frac{4\pi}{3} = -\frac{\sqrt{3}}{2}$ .

Worked example 2. Find $\tan\frac{3\pi}{4}$ .

Quadrant II, reference angle $\frac{\pi}{4}$ : the point is $\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ . Then $\tan\frac{3\pi}{4}$ $= \frac{\sqrt{2}/2}{-\sqrt{2}/2}$ $= -1$ .

Angles outside $[0, 2\pi)$ first move by whole turns: $\frac{11\pi}{4} = 2\pi + \frac{3\pi}{4}$ has the same terminal ray as $\frac{3\pi}{4}$ , so the same values.

Circles of any radius

On a circle of radius $r$ centered at the origin, the point at angle $\theta$ is the unit-circle point scaled by $r$ :

$(x, y)$ $= (r\cos\theta, r\sin\theta)$ .

Worked example 3. A robot arm 2 meters long pivots at the origin. Its tip starts on the positive $x$ -axis and turns counterclockwise through $\frac{7\pi}{6}$ . Where is the tip?

Quadrant III, reference angle $\frac{\pi}{6}$ : $\cos\frac{7\pi}{6} = -\frac{\sqrt{3}}{2}$ and $\sin\frac{7\pi}{6} = -\frac{1}{2}$ .

Scale by 2: the tip is at $\left(2 \cdot \left(-\frac{\sqrt{3}}{2}\right), 2 \cdot \left(-\frac{1}{2}\right)\right)$ $= (-\sqrt{3}, -1)$ .

These exact values are the raw material for the graphs in the next topic and for every no-calculator model in the rest of the unit, so practice until they come without thinking.