Topics
6. Trigonometric Functions
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United States of America
Grades 9-12

6.02 Evaluating trigonometric functions

Lesson
Practice

Introduction

In Geometry, we learned how to use trigonometric ratios to find missing sides and angles of right triangles. We will use those tools along with angle measures in standard form on the coordinate plane from  6.01 Angles and radian measure  to make connections to special triangles and extend the domain of the unit circle beyond 2 \pi revolutions in order to evaluate trigonometric functions with exact ratios.

Trigonometric ratios on the coordinate plane

In lesson  8.03 Trigonometric ratios  , we learned that a trigonometric ratio is a relationship between an acute angle and a pair of sides in a right triangle. We can also find the trigonometric ratio for any angle on the xy-plane using the coordinate of a point on the terminal side of an angle.

The image below shows a point P \left(x, y \right) with a terminal side length r.

A point P at (x, y) plotted on a first quadrant coordinate plane without numbers. A segment with length r is drawn from the origin to P. A vertical segment with length y is drawn from P to a point on the positive x axis. A horizontal segment with length x is drawn from the origin to a point along the positive x axis, and directly below point P. The segment with length r makes an angle labeled theta with respect to the positive x axis.

Once we place trigonometric ratios on the coordinate plane, we can now define trigonometric functions with respect to the acute reference angle \theta as:\begin{aligned} \text{The sine ratio: } \sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}} & = \dfrac{y}{r} \\\\ \text{The cosine ratio: } \cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} & = \dfrac{x}{r} \\\\ \text{The tangent ratio: } \tan \theta = \dfrac{\text{opposite}}{\text{adjacent}} & = \dfrac{y}{x} \end{aligned}

In lesson  6.01 Angles and radian measure  , we examined the measures of angles in standard position with a vertex at \left( 0,0 \right) on the coordinate plane.

Counterclockwise rotations of the terminal side of the angle represent positive angle measures, while clockwise rotations represent negative angle measures.

Angles may be measured in degrees or radians, and may be rational or irrational numbers. Angles larger than 360 \degree or 2 \pi result from continuing to rotate around the circle in multiple rotations. Thus, we can extend the definition of trigonometric functions to include values for all angles \theta in the set of real numbers.

Exploration

Explore the applet by dragging the triangle and checking the box.

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  1. What do you notice about the trigonometric ratio of an angle and its reference angle?
  2. What do you notice about the signs of the ratios in different quadrants if you were to simplify them?

Previously, we defined trigonometric ratios as the lengths of the sides of a right triangle around a reference angle. The ratios were always a measure of distance, and therefore positive.

Now, we define the trigonometric functions in relation to a point on the coordinate plane that forms a right triangle with the x-axis. The terminal side length, r, will still be a distance, and therefore positive. However, since either of the coordinates of a point \left( x, y \right) may be positive or negative, any of the trigonometric ratios may also end up being positive or negative.

There are also three reciprocal trigonometric ratios that we will introduce as trigonometric identities.

Trigonometric identity

A trigonometric equation that is true for all values of the variable for which all expressions in the equation are defined

\begin{aligned} \text{The cosecant ratio: } \csc \theta = \dfrac{1}{\sin \theta} = \dfrac{\text{hypotenuse}}{\text{opposite}} & = \dfrac{r}{y} \\\\ \text{The secant ratio: } \sec \theta = \dfrac{1}{\cos \theta} = \dfrac{\text{hypotenuse}}{\text{adjacent}} & = \dfrac{r}{x} \\\\ \text{The cotangent ratio: } \cot \theta = \dfrac{1}{\tan \theta } = \dfrac{\text{adjacent}}{\text{opposite}} & = \dfrac{x}{y} \end{aligned}

These identities, \csc \theta, \sec \theta, and \cot \theta, are also considered trigonometric functions.

Other identities include:

Tangent identity \qquad \tan \theta = \dfrac{\sin \theta}{\cos \theta} \qquad \qquad Cotangent identity \qquad \cot \theta = \dfrac{\cos \theta}{\sin \theta} \qquad

Examples

Example 1

Consider the angle, \theta= 315 \degree, formed by the positive x-axis and terminal side.

a

Evaluate the six trigonometric functions for \theta. Round your answer to two decimal places.

Worked Solution
Create a strategy

Use a calculator with the mode set to degrees to calculate the six trigonometric functions at 315 \degree.

Apply the idea

\sin 315 \degree = -0.71

\cos 315 \degree = 0.71

\tan 315 \degree = -1

\csc 315 \degree = -1.41

\sec 315 \degree = 1.41

\cot 315 \degree = -1

b

Find the reference angle, \alpha, formed by the terminal side and the x-axis. Evaluate the six trigonometric functions for \alpha and compare your results to part (a).

Worked Solution
Create a strategy

If we draw the angle 315 \degree, we can see that the reference angle is in the fourth quadrant. The difference between 315 \degree and a full circle at 360 \degree is 45 \degree. Hence, the reference angle, \alpha, is 45 \degree.

-1
1
x
-1
1
y
Apply the idea

\sin \left( 45 \degree \right) = 0.71

\cos \left( 45 \degree \right) = 0.71

\tan \left( 45 \degree \right) = 1

\csc \left( 45 \degree \right) = 1.41

\sec \left( 45 \degree \right) = 1.41

\cot \left( 45 \degree \right) = 1

The value of each trigonometric function for the reference angle of 45 \degree is the same as the absolute value of each trigonometric function for the original angle of 315 \degree.

Example 2

The point on the following graph has coordinates \left( -7, -24 \right).

-20
-10
10
20
x
-20
-10
10
20
y
a

Find r, the distance from the point to the origin.

Worked Solution
Create a strategy

Use the distance formula to calculate the length of r.

Apply the idea
\displaystyle d\displaystyle =\displaystyle \sqrt{ \left ( x_2 - x_1 \right)^2 + \left( y_2 - y_1 \right)^2}Distance formula
\displaystyle r\displaystyle =\displaystyle \sqrt{ \left( -7 - 0 \right)^2 + \left( -24 - 0 \right)^2}Substitution
\displaystyle r\displaystyle =\displaystyle \sqrt{49 + 576}Evaluate the subtraction and exponents
\displaystyle r\displaystyle =\displaystyle 25Evaluate the addition and square root
Reflect and check

We could draw a right triangle from the point to the x-axis. The length of r could be calculated with the Pythagorean theorem, which is equivalent to the distance formula.

-20
-10
10
20
x
-20
-10
10
20
y
b

Evaluate the six trigonometric functions for \theta. Leave your answers as a ratio.

Worked Solution
Create a strategy

Use \left(-7, -24 \right) and r=25 to write the six trigonometric ratios.

Apply the idea

\sin \theta = \dfrac{-24}{25}

\cos \theta = \dfrac{-7}{25}

\tan \theta = \dfrac{-24}{-7} = \dfrac{24}{7}

\csc \theta = \dfrac{25}{-24}

\sec \theta = \dfrac{25}{-7}

\cot \theta = \dfrac{-7}{-24} = \dfrac{7}{24}

c

Describe any relationships you notice between the ratios in part (b).

Worked Solution
Apply the idea

The reciprocal ratios flip the ratio of a given trigonometric function. For instance:

\sin \theta = \dfrac{-24}{25} and \csc \theta = \dfrac{1}{\sin \theta} = \dfrac{25}{-24}

\cos \theta = \dfrac{-7}{25} and \sec \theta = \dfrac{1}{\cos \theta} = \dfrac{25}{-7}

\tan \theta = \dfrac{24}{7} and \cot \theta = \dfrac{1}{\tan \theta} = \dfrac{7}{24}

Reflect and check

Note that since the x- and y-coordinates of the point we use are negative, we use the negative values to represent the side lengths of the right triangle's legs, but the hypotenuse remains a positive measure of distance.

The trigonometric functions involving only the legs for the ratio, which are negative numbers based on the point, will result in a positive value.

The trigonometric functions involving the hypotenuse and a leg in the ratio will result in a negative value.

Idea summary

Trigonometric ratios may be formed for any angle on the xy-plane by using the coordinates of a point \left(x, y \right) on the terminal side of the angle \theta, with a domain that includes all real numbers.

Evaluating these trigonometric functions may result in positive or negative values for f \left( \theta \right), depending on the quadrant where \left(x, y \right) is located.

Evaluating trigonometric functions with exact ratios

Exploration

Consider the special triangles drawn in the unit circles shown below.

Two unit circles. On the left unit circle, a radius of length 1 is drawn from the origin to a point on the circle in the first quadrant. A vertical segment is drawn from the point to a point on the positive x axis. A horizontal segment is drawn from the origin to the point along the x axis, directly below the point on the circle. The horizontal and vertical segments form a right angle. The angle formed by the radius and the positive x axis is labeled theta equals 30 degrees. The angle formed by the radius and vertical segment has a measure of 60 degrees. On the right unit circle, a radius of length 1 is drawn from the origin to a point on the circle in the first quadrant. A vertical segment is drawn from the point to a point on the positive x axis. A horizontal segment is drawn from the origin to the point along the x axis, directly below the point on the circle. The horizontal and vertical segments form a right angle. The angle formed by the radius and the positive x axis is labeled theta equals 45 degrees. The angle formed by the radius and vertical segment has a measure of 45 degrees.
  1. Calculate the legs of each triangle, keeping your answer in fraction form.
  2. 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

\tan \theta = \dfrac{y}{x} = \dfrac{\sin \theta}{\cos \theta}

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

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

\cot \theta = \dfrac{x}{y} = \dfrac{ \cos \theta}{\sin \theta}

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 3

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.
a

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
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.
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.
b

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).

c

Fill out the ordered pair for \dfrac{7 \pi}{6} 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
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.
d

Fill out the ordered pair for \dfrac{5 \pi}{3} 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
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 4

Find the exact value of the six trigonometric 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}

\tan \dfrac{2 \pi}{3} = \dfrac{\sin \frac{2 \pi}{3}}{\cos \frac{2 \pi}{3}} = - \sqrt{3}

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

\csc \dfrac{2 \pi}{3}= \dfrac{1}{\sin \frac{2 \pi}{3}} = \dfrac{2}{\sqrt{3}} = \dfrac{2 \sqrt{3}}{3}

\cot \dfrac{2 \pi}{3} =\dfrac{ \cos \frac{2 \pi}{3}}{\sin \frac{2 \pi}{3}} = - \dfrac{1}{\sqrt{3}}=- \dfrac{\sqrt{3}}{3}

Reflect and check

When we know the relationship between sine and cosine and the identities, we can use the values of sine and cosine to derive the exact ratio of the remaining trigonometric functions.

Example 5

Find the exact value of the expression. \sin \dfrac{9 \pi}{4} \left( \tan \dfrac{5 \pi}{6} + \cos \dfrac{5 \pi}{6} \right)

Worked Solution
Create a strategy

Determine the reference angle for \dfrac{9 \pi}{4} and \dfrac{5 \pi}{6}, then use the exact values of the trigonometric functions at the angles and substitute them into the expression.

Apply the idea

\dfrac{9 \pi}{4} shares a terminal side with \dfrac{\pi}{4} in quadrant 1, and the reference angle for \dfrac{5 \pi}{6} which is located in quadrant 2 is \dfrac{\pi}{6}.

\sin \dfrac{9 \pi}{4} = \dfrac{\sqrt{2}}{2}

\tan \dfrac{5 \pi}{6} = \dfrac{ \frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\dfrac{1}{\sqrt{3}} = -\dfrac{\sqrt{3}}{3}

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

\displaystyle \sin \dfrac{9 \pi}{4} \left( \tan \dfrac{5 \pi}{6} + \cos \dfrac{5 \pi}{6} \right)\displaystyle =\displaystyle \dfrac{\sqrt{2}}{2} \left( -\dfrac{\sqrt{3}}{3} + \left( -\dfrac{\sqrt{3}}{2} \right) \right)Substitution
\displaystyle =\displaystyle \dfrac{\sqrt{2}}{2} \left( \dfrac{ -5 \sqrt{3}}{6} \right)Definition of adding rational numbers
\displaystyle =\displaystyle -\dfrac{5 \sqrt{6}}{12}Evaluate the multiplication
Reflect and check

We could also find the exact value of the expression by distributing instead of adding the rational numbers first:

\displaystyle \sin \dfrac{9 \pi}{4} \left( \tan \dfrac{5 \pi}{6} + \cos \dfrac{5 \pi}{6} \right)\displaystyle =\displaystyle \dfrac{\sqrt{2}}{2} \left( -\dfrac{\sqrt{3}}{3} + \left( -\dfrac{\sqrt{3}}{2} \right) \right)Substitution
\displaystyle =\displaystyle -\dfrac{\sqrt{6 }}{6} - \dfrac{ \sqrt{6}}{4}Distribution
\displaystyle =\displaystyle -\dfrac{2 \sqrt{6 }}{12} - \dfrac{ 3\sqrt{6}}{12}Determine a common denominator
\displaystyle =\displaystyle -\dfrac{5 \sqrt{6}}{12}Definition of subtracting rational numbers
Idea summary

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

Two unit circles. On the left unit circle, a radius of length 1 is drawn from the origin to a point on the circle in the first quadrant. A vertical segment is drawn from the point to a point on the positive x axis. A horizontal segment is drawn from the origin to the point along the x axis, directly below the point on the circle. The horizontal and vertical segments form a right angle. The angle formed by the radius and the positive x axis is labeled theta equals 30 degrees. The angle formed by the radius and vertical segment has a measure of 60 degrees. On the right unit circle, a radius of length 1 is drawn from the origin to a point on the circle in the first quadrant. A vertical segment is drawn from the point to a point on the positive x axis. A horizontal segment is drawn from the origin to the point along the x axis, directly below the point on the circle. The horizontal and vertical segments form a right angle. The angle formed by the radius and the positive x axis is labeled theta equals 45 degrees. The angle formed by the radius and vertical segment has a measure of 45 degrees.

Trigonometric functions and the unit circle

Exploration

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

Loading interactive...
  1. In which quadrants are each of sine, cosine and tangent positive?
  2. For each quadrant, list out the ratios which are positive there.
  3. If we know where sine, cosine and tangent are positive/negative, what does this tell us about cosecant, secant and cotangent?
  4. What do you notice about the ratios of angles with the same reference angle? Such as \sin \dfrac{\pi}{6}, \sin \dfrac{5 \pi}{6}, \sin \dfrac{7 \pi}{6}, and \sin \dfrac{ 11 \pi}{6}.

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 6

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 ratio0
a

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 ratio00.50.710.8710.870.710.50
b

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}.

c

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.

d

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 7

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 ratio1
a

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 ratio10.870.710.50-0.5-0.71-0.87-1
b

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.

c

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.

d

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.

Outcomes

F.IF.B.5

Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes.

F.IF.B.6

Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph.

F.TF.A.2

Explain how the unit circle in the coordinate plane enables the extension of trigonometric functions to all real numbers, interpreted as radian measures of angles traversed counterclockwise around the unit circle.

F.TF.A.3 (+)

Use special triangles to determine geometrically the values of sine, cosine, tangent for π/3, π/4 and π/6, and use the unit circle to express the values of sine, cosine, and tangent for x, π x, and 2π - x in terms of their values for x, where x is any real number.

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