3.3 - Sine and Cosine Function Values

Overview

Topic 3.2 defined sine and cosine with a circle. This topic uses those definitions to get actual values. First you'll find the coordinates of any point on a circle centered at the origin. Then you'll use two special right triangles to find exact values of sine and cosine for the angles that show up again and again on the AP exam: multiples of

\frac{\pi}{6}

and

\frac{\pi}{4}

Coordinates of a point on a circle

From Topic 3.2, if the terminal ray of

\theta

meets a circle of radius

r

centered at the origin at the point

P(x, y)

, then

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

Multiply both equations by

r

to solve for the coordinates:

x = r\cos\theta \qquad y = r\sin\theta

So the point is

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

Think of it as the unit-circle point

(\cos\theta, \sin\theta)

stretched out by a factor of

r

. This shows up any time something moves around a circle: a seat on a Ferris wheel, the tip of a clock hand, a point on a spinning wheel.

Example 1: A point on a circle, with a calculator

A circle centered at the origin has radius 10. Find the coordinates of the point where the terminal ray of

\theta = 2

radians meets the circle.

P = (10\cos 2,\ 10\sin 2) \approx \big(10(-0.41615),\ 10(0.90930)\big) \approx (-4.161,\ 9.093)

Check the quadrant. 2 radians is between

\frac{\pi}{2} \approx 1.571

and

\pi \approx 3.142

, so the point should be in Quadrant II, with a negative

x

-coordinate and a positive

y

-coordinate. It is.

Answer:

P \approx (-4.161, 9.093)

Exact values from special right triangles

For most angles, sine and cosine are messy decimals that you'd get from a calculator. But for angles that are multiples of

\frac{\pi}{6}

(

30^\circ

) or

\frac{\pi}{4}

(

45^\circ

), geometry gives exact values. Two triangles do all the work. Put each one inside the unit circle so its hypotenuse is a radius, with length 1.

The isosceles right triangle: $\frac{\pi}{4}$

An angle of

\frac{\pi}{4}

makes an isosceles right triangle. Both acute angles are

\frac{\pi}{4}

, so both legs have the same length, call it

a

. With a hypotenuse of 1, the Pythagorean theorem gives

a^2 + a^2 = 1 \;\Rightarrow\; 2a^2 = 1 \;\Rightarrow\; a = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}

\cos\frac{\pi}{4} = \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} \approx 0.707

Half of an equilateral triangle: $\frac{\pi}{6}$ and $\frac{\pi}{3}$

Start with an equilateral triangle whose sides are all 1 and whose angles are all

\frac{\pi}{3}

. Cut it in half with a line from one vertex straight down to the middle of the opposite side. Each half is a right triangle with a hypotenuse of 1, a short leg of

\frac{1}{2}

(half of a side), and acute angles of

\frac{\pi}{6}

and

\frac{\pi}{3}

. The long leg is

\sqrt{1^2 - \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}

The short leg is always across from the smaller angle. So:

At $\theta = \frac{\pi}{6}$ , the vertical leg is the short one: $P = \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ .

At $\theta = \frac{\pi}{3}$ , the vertical leg is the long one: $P = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ .

Together with the points on the axes, which come straight from the unit circle, here are the Quadrant I values:

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

A pattern that helps you remember: as

\theta

goes from

0

\frac{\pi}{2}

, sine climbs through

0, \frac{1}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, 1

, and cosine runs through the same list backward.

Angles in other quadrants: Reference angles

For an angle like

\frac{5\pi}{6}

that doesn't end in Quadrant I, use its reference angle: the acute angle between the terminal ray and the

x

-axis. The triangle made with the reference angle is one of the special triangles, so the coordinates have the same sizes as in Quadrant I. Only the signs change, and the quadrant tells you which signs to use.

If needed, add or subtract $2\pi$ until you have a coterminal angle between $0$ and $2\pi$ .

Find the quadrant and the reference angle.

Look up the Quadrant I values for the reference angle.

Attach signs: $x$ (cosine) is negative in Quadrants II and III, and $y$ (sine) is negative in Quadrants III and IV.

Terminal ray in	Reference angle
Quadrant II	$\pi - \theta$
Quadrant III	$\theta - \pi$
Quadrant IV	$2\pi - \theta$

Doing this for every multiple of

\frac{\pi}{6}

and

\frac{\pi}{4}

gives the full unit circle:

You don't need to memorize this picture point by point. If you know the three Quadrant I points and the sign rules, you can rebuild any part of it in a few seconds, and that's less likely to go wrong under exam pressure.

Example 2: An angle in Quadrant II

Find the exact values of

\cos\frac{5\pi}{6}

and

\sin\frac{5\pi}{6}

Step 1: Quadrant and reference angle.

\frac{5\pi}{6}

is between

\frac{\pi}{2}

and

\pi

, so it's in Quadrant II. The reference angle is

\pi - \frac{5\pi}{6} = \frac{\pi}{6}

Step 2: Quadrant I values. For

\frac{\pi}{6}

, the point is

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

Step 3: Signs. In Quadrant II,

x

is negative and

y

is positive.

The point for

\frac{5\pi}{6}

is the mirror image of the point for

\frac{\pi}{6}

(gray) across the

y

-axis. Same triangle, opposite side.

Answer:

\cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}

and

\sin\frac{5\pi}{6} = \frac{1}{2}

Example 3: A negative angle

Find the exact values of

\cos\left(-\frac{3\pi}{4}\right)

\sin\left(-\frac{3\pi}{4}\right)

, and

\tan\left(-\frac{3\pi}{4}\right)

Step 1: Coterminal angle.

-\frac{3\pi}{4} + 2\pi = \frac{5\pi}{4}

, which is between

\pi

and

\frac{3\pi}{2}

, so it's in Quadrant III.

Step 2: Reference angle.

\frac{5\pi}{4} - \pi = \frac{\pi}{4}

Step 3: Signs. Both coordinates are negative in Quadrant III.

\cos\left(-\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2} \qquad \sin\left(-\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}

\tan\left(-\frac{3\pi}{4}\right) = \frac{-\sqrt{2}/2}{-\sqrt{2}/2} = 1

Answer:

\cos\left(-\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}

\sin\left(-\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}

, and

\tan\left(-\frac{3\pi}{4}\right) = 1

Example 4: An angle bigger than $2\pi$

Find the exact values of

\cos\frac{11\pi}{3}

and

\sin\frac{11\pi}{3}

Step 1: Coterminal angle.

\frac{11\pi}{3} - 2\pi = \frac{11\pi}{3} - \frac{6\pi}{3} = \frac{5\pi}{3}

, which is between

\frac{3\pi}{2}

and

2\pi

, so it's in Quadrant IV.

Step 2: Reference angle.

2\pi - \frac{5\pi}{3} = \frac{\pi}{3}

, and the Quadrant I point for

\frac{\pi}{3}

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

Step 3: Signs. In Quadrant IV,

x

is positive and

y

is negative.

Answer:

\cos\frac{11\pi}{3} = \frac{1}{2}

and

\sin\frac{11\pi}{3} = -\frac{\sqrt{3}}{2}

Example 5: A spinning wheel (AP-style)

A bicycle sits on a repair stand, and its back wheel, with a radius of 35 cm, is free to spin. Put the center of the wheel at the origin. A reflector on the rim starts at

(35, 0)

, level with the center, and the wheel turns counterclockwise. The center of the wheel is 60 cm above the floor.

(a) After the wheel turns through

\frac{5\pi}{6}

radians, find the exact coordinates of the reflector. (b) How high above the floor is the reflector then? (c) The wheel keeps turning until it has turned through

\frac{4\pi}{3}

radians in total. How high above the floor is the reflector now?

(a) Use

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

with

r = 35

and the values from Example 2:

P = \left(35\cos\frac{5\pi}{6},\ 35\sin\frac{5\pi}{6}\right) = \left(35 \cdot \left(-\frac{\sqrt{3}}{2}\right),\ 35 \cdot \frac{1}{2}\right) = \left(-\frac{35\sqrt{3}}{2},\ \frac{35}{2}\right)

That's about

(-30.311, 17.5)

: left of the center and above it, as expected for Quadrant II.

(b) The

y

-coordinate is the reflector's height above the center, so its height above the floor is

60 + \frac{35}{2} = 60 + 17.5 = 77.5 \text{ cm}

(c)

\frac{4\pi}{3}

is in Quadrant III with reference angle

\frac{4\pi}{3} - \pi = \frac{\pi}{3}

, so

\sin\frac{4\pi}{3} = -\frac{\sqrt{3}}{2}

. The reflector is below the center:

y = 35\sin\frac{4\pi}{3} = -\frac{35\sqrt{3}}{2} \approx -30.311

\text{height} = 60 - \frac{35\sqrt{3}}{2} \approx 29.689 \text{ cm}

Answer: (a)

\left(-\frac{35\sqrt{3}}{2}, \frac{35}{2}\right)

. (b) 77.5 cm above the floor. (c) About 29.689 cm above the floor, which is about 30.311 cm below the center of the wheel.

Common mistakes

Swapping two special values. The values for $\frac{\pi}{6}$ and $\frac{\pi}{3}$ are easy to mix up, so picture the point. At $\frac{\pi}{6}$ the point is low and far to the right, so its $y$ -coordinate (sine) is the small one, $\frac{1}{2}$ .

Forgetting signs outside Quadrant I. The triangle gives the sizes, and the quadrant gives the signs.

Measuring the reference angle to the y-axis. The reference angle is always measured to the $x$ -axis.

Forgetting the radius. On a circle of radius $r$ , the point is $(r\cos\theta, r\sin\theta)$ , not $(\cos\theta, \sin\theta)$ .

Not recognizing equivalent forms. $\frac{1}{\sqrt{2}}$ and $\frac{\sqrt{2}}{2}$ are the same number. Answer choices usually use $\frac{\sqrt{2}}{2}$ .

Back Next

3.3 - Sine and Cosine Function Values

Overview

Coordinates of a point on a circle

Example 1: A point on a circle, with a calculator

Exact values from special right triangles

The isosceles right triangle: π4\frac{\pi}{4}4π​

Half of an equilateral triangle: π6\frac{\pi}{6}6π​ and π3\frac{\pi}{3}3π​

Angles in other quadrants: Reference angles

Example 2: An angle in Quadrant II

Example 3: A negative angle

Example 4: An angle bigger than 2π2\pi2π

Example 5: A spinning wheel (AP-style)

Common mistakes

The isosceles right triangle: $\frac{\pi}{4}$

Half of an equilateral triangle: $\frac{\pi}{6}$ and $\frac{\pi}{3}$

Example 4: An angle bigger than $2\pi$