Calculate the legs of each triangle, keeping your answer in fraction form.
Write the ordered pair that represents the endpoint of each radius on each unit circle.

Now we will consider the exact ratios of trigonometric functions when a point, P, is on the unit circle, where the center is at the origin and the radius is 1.

A unit circle plotted on a four quadrant coordinate plane. A point P at (x, y) is on the circle on the first quadrant. A radius of length 1 is drawn from the origin to P. A dashed vertical segment of length y is drawn from P to a point on the x axis. A horizontal segment of length x is drawn from the origin to a point along the x axis and directly below P. The horizontal and dashed vertical segment forms a right angle. The interior angle formed by the radius and the positive x axis is labeled theta.

Consider a point P with coordinates \left(x, y \right):

\cos \theta = \dfrac{x}{1} = x

\sin \theta = \dfrac{y}{1}=y

Therefore, the ordered pair \left(x, y \right) on the coordinate plane can be represented by \left(\cos \theta, \sin \theta \right) when evaluating trigonometric functions on the unit circle.

Special triangles will be used to help us determine exact ratios. Recall that special triangles will always have side lengths that are in proportion. A 45 \degree- 45 \degree- 90\degree triangle has side lengths with a ratio of 1:1:\sqrt{2}. A 30 \degree- 60 \degree- 90\degree triangle will have side lengths of proportion 1: \sqrt{3}:2.

Right triangle A B C with right angle C. The legs of the triangle have a length of s, and the hypotenuse has a length of s square root of 2. Angles A and B have a measure of 45 degrees.

45 \degree- 45 \degree- 90\degree triangle

Right triangle A B C with right angle C. Leg B C has length of s, leg A C has length of s square root of 3, and hypotenuse A B has a length of 2 s. Angle A has a measure of 30 degrees, and angle B has a measure of 60 degrees.

30 \degree- 60 \degree- 90\degree triangle

Examples

Example 1

Consider the special triangles on each unit circle shown below:

Three unit circles. The left unit circle has a point at (square root of 3 over 2, 1 half). A radius of length 1 is drawn from the origin to the point. A vertical segment of length 1 half is drawn from the point to a point on the x axis. A horizontal segment of length square root of 3 over 2 is drawn from the origin to a point along the x axis and directly below the point on the unit circle. The horizontal and dashed vertical segments form a right angle. The interior angle formed by the radius and the positive x axis measures 30 degrees, and the angle formed by the by the radius and the vertical segment measures 60 degrees. The middle unit circle has a point at (square root of 2 over 2, square root of 2 over 2). A radius of length 1 is drawn from the origin to the point. A vertical segment of length square root of 2 over 2 is drawn from the point to a point on the x axis. A horizontal segment of length square root of 2 over 2 is drawn from the origin to a point along the x axis and directly below the point on the unit circle. The horizontal and dashed vertical segments form a right angle. The interior angle formed by the radius and the positive x axis measures 45 degrees, and the angle formed by the by the radius and the vertical segment measures 45 degrees. The right unit circle has a point at (1 half, square root of 3 over 2). A radius of length 1 is drawn from the origin to the point. A vertical segment of length square root of 3 over 2 is drawn from the point to a point on the x axis. A horizontal segment of length 1 half is drawn from the origin to a point along the x axis and directly below the point on the unit circle. The horizontal and dashed vertical segments form a right angle. The interior angle formed by the radius and the positive x axis measures 60 degrees, and the angle formed by the by the radius and the vertical segment measures 30 degrees.

Fill out the missing ordered pairs on the unit circle in the first quadrant.

The unit circle plotted in a first quadrant coordinate plane and with the special angles labeled. Starting from the positive x axis then moving counterclockwise, the special angles are: 0, pi over 6, pi over 4, pi over 3, and pi over 2. Speak to your teacher for more details.

Worked Solution

Create a strategy

Use the coordinates using the reference angles of special triangles from the unit circles provided to fill out the ordered pairs in the first quadrant.

Apply the idea

Reflect and check

Note that the ratio of the side lengths for a special 45 \degree- 45 \degree- 90\degree triangle is 1:1:\sqrt{2}, so the hypotenuse is \sqrt{2} times the length of one leg. Since the hypotenuse of the 45 \degree- 45 \degree- 90\degree triangle on the unit circle is 1 unit, then its side lengths must each equal \dfrac{1}{\sqrt{2}}. In the given special triangle, the fraction is written with a rationalized denominator, so \dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}. We will use the fraction with a rationalized denominator for the special triangle on the unit circle.

Two right triangles. The left triangle has leg lengths of 1 over square of 2 and hypotenuse length of 1. The angle opposite one of the 1 over square of 2 leg has a measure of 45 degrees. The right triangle has leg lengths of square root of 2 over 2 and hypotenuse length of 1. The angle opposite one of the square root of 2 over 2 leg has a measure of 45 degrees.

Fill out the ordered pair for \dfrac{3 \pi}{4} on the unit circle:

The unit circle with the special angles labeled. Starting from the positive x axis then moving counterclockwise, the special angles are: 0, pi over 6, pi over 4, pi over 3, pi over 2, 2 pi over 3, 3 pi over 4, 5 pi over 6, pi, 7 pi over 6, 5 pi over 4, 4 pi over 3, 3 pi over 2, 5 pi over 3, 7 pi over 4, and 11 pi over 6. Some parts of the unit circle are missing. Speak to your teacher for more details.

Worked Solution

Create a strategy

Determine the reference angle of \dfrac{3 \pi}{4} on the unit circle, then use the corresponding special triangle from the first quadrant to write the ordered pair. Since \dfrac{3 \pi}{4} is in the second quadrant, we will need to use the appropriate sign for the x- and y-coordinates.

Apply the idea

Since the reference angle of \dfrac{3 \pi}{4} is \dfrac{\pi}{4}, we can use the side lengths of the special 45 \degree- 45 \degree- 90\degree triangle in the first quadrant. Its coordinates in the first quadrant are \left( \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2} \right). \dfrac{3 \pi}{4} is located in the second quadrant, where x is negative, while y remains positive. Thus, the ordered pair for \dfrac{3 \pi}{4} is \left( -\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2} \right).

Fill out the ordered pair for \dfrac{7 \pi}{6} on the unit circle:

Worked Solution

Apply the idea

The angle \dfrac{7 \pi}{6} is in the third quadrant, where both coordinates are negative and the reference angle is \dfrac{\pi}{6}, so the ordered pair is \left( -\dfrac{\sqrt{3}}{2} , -\dfrac{1}{2} \right) .

Reflect and check

We could also choose to draw the special triangle on the unit circle at the given angle, and label it with the appropriate signs for x and y according to its coordinates from the first quadrant.

The unit circle on the third quadrant with 3 unlabeled points on the circle. Segments are drawn from the origin to each of the points. Speak to your teacher for more details.

Fill out the ordered pair for \dfrac{5 \pi}{3} on the unit circle:

Worked Solution

Apply the idea

The angle \dfrac{5 \pi}{3} is in the fourth quadrant, where the x-coordinate is positive and the y-coordinate is negative. Since the reference angle is \dfrac{\pi}{3}, the ordered pair at \dfrac{5 \pi}{3} is \left( \dfrac{1}{2}, -\dfrac{\sqrt{3}}{2} \right).

Reflect and check

Knowing the special triangles on the first quadrant of the unit circle and using symmetry to recall angles in the other quadrants will allow us to evaluate exact trigonometric functions efficiently.

Example 2

Find the exact value of the sine and cosine functions when \theta = \dfrac{2 \pi}{3}.

Worked Solution

Create a strategy

\dfrac{2 \pi}{3} is located in quadrant 2, where x or \cos \theta is negative and y or \sin \theta is positive. Then use the reference angle \dfrac{\pi}{3} and the x- and y-coordinates from the first quadrant to evaluate the trigonometric functions.

Apply the idea

\cos \dfrac{2 \pi}{3} = -\dfrac{1}{2}

\sin \dfrac{2 \pi}{3} = \dfrac{\sqrt{3}}{2}

Idea summary

The special triangles on the first quadrant of the unit circle are a useful tool for evaluating trigonometric functions with exact ratios:

Sine and cosine functions and the unit circle

Exploration

Explore the applet by dragging the slider and checking the boxes.

Loading interactive...

In which quadrants are each of sine and cosine positive?
For each quadrant, list out the ratios which are positive there.

As the angle \theta changes, its measure growing larger as it rotates counterclockwise around the origin, \sin \theta represents the change in the y-coordinate, or the height of the right triangle. At the same time, \cos \theta represents the change in the x-coordinate, or the length of the base leg of the triangle. The radius of the circle or the hypotenuse of the right triangle will remain 1.

Examples

Example 3

Consider the table of values for the sine function at different angles on the unit circle.

	\sin 0	\sin \frac{\pi}{6}	\sin \frac{\pi}{4}	\sin \frac{\pi}{3}	\sin \frac{\pi}{2}	\sin \frac{2\pi}{3}	\sin \frac{3\pi}{4}	\sin \frac{5\pi}{6}	\sin \pi
Evaluated ratio	0

Complete the table of values for the sine ratios at the given angles. Round your answers to two decimal places.

Worked Solution

Create a strategy

Find the y-coordinate at each angle on the unit circle, and simplify the coordinate using a calculator.

Apply the idea

	\sin 0	\sin \frac{\pi}{6}	\sin \frac{\pi}{4}	\sin \frac{\pi}{3}	\sin \frac{\pi}{2}	\sin \frac{2\pi}{3}	\sin \frac{3\pi}{4}	\sin \frac{5\pi}{6}	\sin \pi
Evaluated ratio	0	0.5	0.71	0.87	1	0.87	0.71	0.5	0

Describe what happens to \sin \theta as \theta approaches \dfrac{\pi}{2} from 0.

Worked Solution

Create a strategy

Using the table from part (a), review the value of the ratio in the first five entries of the table.

Apply the idea

\sin \theta starts at 0, then increases at a decreasing rate until it gets to 1 at \dfrac{\pi}{2}.

Describe what happens to \sin \theta as \theta approaches \pi from \dfrac{\pi}{2}.

Worked Solution

Apply the idea

\sin \theta has a value of 1 at \dfrac{\pi}{2}, then decreases at an increasing rate until it gets to 0 at \pi.

Reflect and check

A symmetrical pattern occurs from 0 to \pi, and since we know that \sin \theta represents the y-coordinates on the unit circle, it makes sense that the values of y move from 0 to 1 and back to 0 as the angles change from \theta = 0 to \theta = \pi.

For what values of \theta will \sin \theta be positive? Negative?

Worked Solution

Create a strategy

We have made the connection between the location of a point on the unit circle and the value of \sin \theta at that point for angles from 0 to \pi. By using the rest of the unit circle, we can see how \sin \theta or the y-coordinates change.

Apply the idea

\sin \theta will be positive for values of \theta between 0 and \pi.

\sin \theta will be negative for values of \theta between \pi and 2\pi.

Reflect and check

Note that angles between 0 and \pi are located in the first and second quadrants, where y-coordinates are always positive, and angles between \pi and 2\pi are located in the third and fourth quadrants, where y-coordinates are always negative.

Example 4

Consider the table of values for the cosine ratio at different angles on the unit circle.

	\cos 0	\cos \frac{\pi}{6}	\cos \frac{\pi}{4}	\cos \frac{\pi}{3}	\cos \frac{\pi}{2}	\cos \frac{2\pi}{3}	\cos \frac{3\pi}{4}	\cos \frac{5\pi}{6}	\cos \pi
Evaluated ratio	1

Complete the table of values for the cosine ratios at the given angles. Round your answers to two decimal places.

Worked Solution

Create a strategy

Find the x-coordinate at each angle on the unit circle, and simplify the coordinate using a calculator.

Apply the idea

	\cos 0	\cos \frac{\pi}{6}	\cos \frac{\pi}{4}	\cos \frac{\pi}{3}	\cos \frac{\pi}{2}	\cos \frac{2\pi}{3}	\cos \frac{3\pi}{4}	\cos \frac{5\pi}{6}	\cos \pi
Evaluated ratio	1	0.87	0.71	0.5	0	-0.5	-0.71	-0.87	-1

Describe what happens to \cos \theta as \theta approaches \dfrac{\pi}{2} from 0.

Worked Solution

Create a strategy

Using the table from part (a), review the value of the ratio in the first five entries of the table.

Apply the idea

\cos \theta starts at 1, then decreases at an increasing rate until it gets to 0 at \dfrac{\pi}{2}.

Reflect and check

This is the opposite pattern of \sin \theta over the given domain.

Describe what happens to \cos \theta as \theta approaches \pi from \dfrac{\pi}{2}.

Worked Solution

Apply the idea

\cos \theta has a value of 0 at \dfrac{\pi}{2}, then decreases at a decreasing rate until it gets to -1 at \pi.

Reflect and check

Since we know that \cos \theta represents the x-coordinates on the unit circle, it makes sense that the values of x move from 1 to 0 and -1 as the angles change from \theta = 0 to \theta = \pi.

For what values of \theta will \cos \theta be positive? Negative?

Worked Solution

Create a strategy

We have made the connection between the location of a point on the unit circle and the value of \cos \theta at that point for angles from 0 to \pi. By using the rest of the unit circle, we can see how \cos \theta or the x-coordinates change.

Apply the idea

\cos \theta will be positive for values of \theta between 0 and \dfrac{\pi}{2}, and then for values of \theta between \dfrac{3 \pi}{2} and 2\pi.

\cos \theta will be negative for values of \theta between \dfrac{\pi}{2} and \dfrac{3 \pi}{2}.

Reflect and check

Note that angles between 0 and \dfrac{\pi}{2} and between \dfrac{3 \pi}{2} and 2 \pi are located in the first and fourth quadrants, where x-coordinates are always positive.

Angles between \dfrac{\pi}{2} and \dfrac{3 \pi}{2} are located in the second and third quadrants, where x-coordinates are always negative.

Idea summary

four quadrant coordinate plane with quadrants 1, 2, 3, and 4 labeled. In quadrant 1, the label positive x, positive y is shown; in quadrant 2, negative x, positive y; in quadrant 3, negative x, negative y; and in quadrant 4, positive x, negative y.

We can determine the sign of the six basic trigonometric functions by relating the point \left( x, y \right) on the unit circle to the trigonometric functions \left ( \cos \theta , \sin \theta \right).

In the quadrants where x is positive or negative, \cos \theta will be positive or negative. In the quadrants where y is positive or negative, \sin \theta is positive or negative.

As we rotate counterclockwise around the unit circle, \sin \theta and \cos \theta increase and decrease between -1 and 1.

3.3 Sine and cosine function values

Introduction

Ideas

Evaluate sine and cosine functions with exact ratios

Exploration

Examples

Example 1

Example 2

Sine and cosine functions and the unit circle

Exploration

Examples

Example 3

Example 4

Outcomes

3.3.A

What is Mathspace

About Mathspace