Sine and Cosine Function Values — AP® Precalculus

Part 1 · Hook

Why this matters

On the exam's no-calculator part, and in every question about periodic models, you need values like

\cos\frac{2\pi}{3}

instantly and exactly. You do not have to memorize sixteen points. Two triangles and one rule about signs produce every one of them.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. A 30-60-90 triangle has hypotenuse $1$ . How long is the leg across from the $30^\circ$ angle?

$\frac{1}{2}$
$\frac{\sqrt{3}}{2}$
$2$
$\frac{\sqrt{2}}{2}$

Show the answer

The sides are $1 : \sqrt{3} : 2$ ; scaled to hypotenuse 1 they are $\frac{1}{2} : \frac{\sqrt{3}}{2} : 1$ . The short leg is across from $30^\circ$ .

Correct: $\frac{1}{2}$ :
$\frac{\sqrt{3}}{2}$ :
$2$ :
$\frac{\sqrt{2}}{2}$ :

2. The terminal ray of $\theta$ meets the unit circle at $(0.6, -0.8)$ . What is $\cos\theta$ ?

$0.6$
$-0.8$
$-\frac{4}{3}$
$0.8$

Show the answer

$\cos\theta$ is the $x$ -coordinate of the point.

Correct: $0.6$ :
$-0.8$ :
$-\frac{4}{3}$ :
$0.8$ :

Part 4 · See it

See it first

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. — Every multiple of $\frac{\pi}{6}$ and $\frac{\pi}{4}$ . The first-quadrant values repeat in every quadrant; only the signs change. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

An angle of $\frac{\pi}{6}$ , $\frac{\pi}{4}$ or $\frac{\pi}{3}$ on the unit circle makes a right triangle with hypotenuse 1it is a special right triangle, so the coordinates are exact: $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ , $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ , $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$
An angle in another quadrant, such as $\frac{5\pi}{6}$ its reference angle, the acute angle between the terminal ray and the $x$ -axis, is $\frac{\pi}{6}$ , so its coordinates have the same sizes
The quadrant decides the signs $\frac{5\pi}{6}$ is in Quadrant II, so $\cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$ and $\sin\frac{5\pi}{6} = \frac{1}{2}$
A circle of radius $r$ instead of 1scale the coordinates by $r$ : the point at angle $\theta$ is $(r\cos\theta, r\sin\theta)$

Part 6 · Key ideas

Key ideas

The special angles are the multiples of $\frac{\pi}{6}$ and $\frac{\pi}{4}$ . Their sines and cosines are $0$ , $\pm\frac{1}{2}$ , $\pm\frac{\sqrt{2}}{2}$ , $\pm\frac{\sqrt{3}}{2}$ or $\pm 1$ .
Method: find the reference angle, take the first-quadrant value, then give it the sign of the quadrant.
For a reference angle of $\frac{\pi}{6}$ or $\frac{\pi}{3}$ , the coordinate that is larger in size is $\frac{\sqrt{3}}{2}$ : it belongs to the axis the ray is closer to.
On a circle of radius $r$ centered at the origin, the point at angle $\theta$ is $(r\cos\theta, r\sin\theta)$ .

Part 7 · Misconception

A common mistake

The wrong idea: $\sin\frac{2\pi}{3} = -\frac{\sqrt{3}}{2}$ , because the angle is past $\frac{\pi}{2}$ , so everything turns negative.

What actually happens: Only the coordinate on the negative side changes sign. $\frac{2\pi}{3}$ is in Quadrant II, left of the $y$ -axis but above the $x$ -axis: $\cos\frac{2\pi}{3} = -\frac{1}{2}$ and $\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

A point on a circle of radius 4

The circle

x^2 + y^2 = 16

is centered at the origin

O

. The point

Q

is on the circle, on the terminal ray of the angle

\frac{2\pi}{3}

in standard position.

1. What is the $x$ -coordinate of $Q$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$x = 4\cos\frac{2\pi}{3}$ . The reference angle is $\frac{\pi}{3}$ and Quadrant II makes cosine negative: $4\left(-\frac{1}{2}\right) = -2$ .

Answer: -2

2. What is the $y$ -coordinate of $Q$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$y$ $= 4\sin\frac{2\pi}{3}$ $= 4 \cdot \frac{\sqrt{3}}{2}$ $= 2\sqrt{3}$ .

Answer: $2\,\sqrt{3}$ ≈ 3.464

3. What is the slope of the segment $OQ$ ?

$-\sqrt{3}$
$-\frac{\sqrt{3}}{3}$
$\sqrt{3}$
$-\frac{1}{2}$

Show the answer

The slope is $\tan\frac{2\pi}{3}$ $= \frac{\sqrt{3}/2}{-1/2}$ $= -\sqrt{3}$ ; with the coordinates, $\frac{2\sqrt{3}}{-2} = -\sqrt{3}$ . The radius does not change the slope.

Correct: $-\sqrt{3}$ : Rise $2\sqrt{3}$ over run $-2$ .
$-\frac{\sqrt{3}}{3}$ : This is $\tan\frac{5\pi}{6}$ : it divides $x$ by $y$ instead of $y$ by $x$ .
$\sqrt{3}$ : The segment rises to the left, so the slope is negative.
$-\frac{1}{2}$ : This is $\cos\frac{2\pi}{3}$ , a coordinate, not the slope of the ray.

4. The point $R$ is on the same circle, on the terminal ray of $\frac{2\pi}{3} + \pi$ . What are the coordinates of $R$ ?

$(2, -2\sqrt{3})$
$(-2, -2\sqrt{3})$
$(2\sqrt{3}, -2)$
$(-2, 2\sqrt{3})$

Show the answer

Half a turn more points the ray the opposite way, so $R$ is diametrically opposite $Q$ : both coordinates change sign, $(2, -2\sqrt{3})$ . Check: $\frac{5\pi}{3}$ is in Quadrant IV with reference angle $\frac{\pi}{3}$ .

Correct: $(2, -2\sqrt{3})$ : Opposite $Q$ through the center: $(2, -2\sqrt{3})$ .
$(-2, -2\sqrt{3})$ : This flips only the $y$ -coordinate, the reflection across the $x$ -axis, which is the angle $\frac{4\pi}{3}$ .
$(2\sqrt{3}, -2)$ : This swaps the sizes; with reference angle $\frac{\pi}{3}$ the larger size belongs to $y$ .
$(-2, 2\sqrt{3})$ : This is $Q$ itself, the point for a full turn $2\pi$ , not half a turn.

5. What is the exact value of $\sin\frac{5\pi}{4}$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Quadrant III, reference angle $\frac{5\pi}{4} - \pi = \frac{\pi}{4}$ : $\sin\frac{5\pi}{4} = -\frac{\sqrt{2}}{2}$ .

Answer: $\frac{-\sqrt{2}}{2}$ ≈ -0.707

6. What is the exact value of $\cos\frac{2\pi}{3}$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Quadrant II, reference angle $\pi - \frac{2\pi}{3} = \frac{\pi}{3}$ : $\cos\frac{2\pi}{3} = -\frac{1}{2}$ .

Answer: $\frac{-1}{2}$ ≈ -0.500

7. A point is on a circle of radius 10 centered at the origin, on the terminal ray of $\frac{\pi}{6}$ . What are its coordinates?

$(5\sqrt{3}, 5)$
$(5, 5\sqrt{3})$
$\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
$(10\sqrt{3}, 10)$

Show the answer

$(r\cos\theta, r\sin\theta)$ $= \left(10 \cdot \frac{\sqrt{3}}{2}, 10 \cdot \frac{1}{2}\right)$ $= (5\sqrt{3}, 5)$ .

Correct: $(5\sqrt{3}, 5)$ : Scale $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ by 10.
$(5, 5\sqrt{3})$ : This is the point at $\frac{\pi}{3}$ : the coordinates are swapped.
$\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ : This is the point on the unit circle; multiply by the radius 10.
$(10\sqrt{3}, 10)$ : This doubles the right answer: $10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3}$ , not $10\sqrt{3}$ .

Part 9 · Summary

Summary

Exact sines and cosines at multiples of

\frac{\pi}{6}

and

\frac{\pi}{4}

come from the 30-60-90 and 45-45-90 triangles. Find the reference angle, use its first-quadrant value, and give it the sign of the quadrant. On a circle of radius

r

, the point at angle

\theta

is

(r\cos\theta, r\sin\theta)

.

Part 10 · Up next

What comes next

Part 11 · Connections