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

6.03 Trigonometric identities

Lesson
Practice

Introduction

The relationships between the reciprocal functions we learned about in lesson  6.02 Evaluating trigonometric functions  make it possible to write trigonometric expressions in equivalent forms. We will use the Pythagorean theorem and trigonometric functions to manipulate identities.

Trigonometric identities

Exploration

Drag the slider to change the radius of the circle. Drag the triangle anywhere on the applet.

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  1. Describe what is happening for each step.
  2. How do the steps relate to the Pythagorean theorem?

The Pythagorean theorem concerning right triangles can be expressed by means of trigonometric ratios.

The following Pythagorean identities can also be used to find trigonometric function values given certain information:\begin{aligned} \left(\sin \theta \right)^2 + \left(\cos \theta \right) ^2 = 1 \quad \text{ or } \quad \sin^2 \theta + \cos^2 \theta = 1 \\\ \left(\tan \theta \right) ^ 2 + 1 = \left(\sec \theta \right) ^ 2 \quad \text{ or } \quad \tan ^2 \theta + 1 = \sec^2 \theta \\\ 1 + \left( \cot \theta \right) ^ 2 = \left( \csc \theta \right) ^2 \quad \text{ or } \quad 1 + \cot^2 \theta = \csc^2 \theta \end{aligned}

Recall the basic trigonometric identities:

\begin{aligned} \sin \theta = \dfrac{1}{\csc \theta} \qquad \cos \theta = \dfrac{1}{\sec \theta} \qquad \tan \theta = \dfrac{1}{\cot \theta} \qquad \tan \theta = \dfrac{ \sin \theta}{\cos \theta} \\\\ \csc \theta= \dfrac{1}{\sin \theta} \qquad \sec \theta = \dfrac{1}{\cos \theta} \qquad \cot \theta = \dfrac{1}{\tan \theta} \qquad \cot \theta = \dfrac{\cos \theta}{\sin \theta} \end{aligned}

Using some combination of the above identities, we can derive the values of all six of the trigonometric functions if we are given the value of one of them. We can also rewrite trigonometric expressions in equivalent but more concise ways.

Examples

Example 1

Given the following diagram, prove the Pythagorean identity \sin ^2 \theta + \cos ^2 \theta = 1.

A right triangle with legs of length a and 2 and hypotenuse of length 3. The angle opposite the side of length 2 is labeled theta.
Worked Solution
Create a strategy

Use the Pythagorean theorem with the given triangle to begin the proof. Then, make connections to the sine and cosine of the reference angle to get to the Pythagorean identity.

Apply the idea
\displaystyle a^2 + b^2\displaystyle =\displaystyle c^2Pythagorean theorem
\displaystyle \dfrac{a^2}{c^2} + \dfrac{b^2}{c^2}\displaystyle =\displaystyle 1Divide both sides of the equation by c^2
\displaystyle \left( \dfrac{a}{c} \right) ^2 + \left( \dfrac{b}{c} \right) ^2\displaystyle =\displaystyle 1Apply the quotient to a power property of exponents
\displaystyle \cos \theta= \dfrac{a}{c} \text{ and } \sin \theta\displaystyle =\displaystyle \dfrac{b}{c}Evaluate the sine and cosine of the reference angle
\displaystyle \left(\cos \theta \right) ^2 + \left( \sin \theta \right)^2\displaystyle =\displaystyle 1Substitution
\displaystyle \sin ^2 \theta + \cos ^2 \theta\displaystyle =\displaystyle 1Commutative property of addition

Example 2

Use \sin^2 \theta + \cos^2 \theta = 1 to prove the Pythagorean identity 1 + \cot^2 \theta = \csc^2 \theta.

Worked Solution
Create a strategy

Use reciprocal identities and the cotangent identity to rewrite the expressions in terms of sine and cosine.

Apply the idea
\displaystyle \sin^2 \theta + \cos^2 \theta \displaystyle =\displaystyle 1Pythagorean identity
\displaystyle \dfrac{\sin ^2 \theta}{\sin^2 \theta} + \dfrac{\cos^2 \theta}{\sin^2 \theta}\displaystyle =\displaystyle \dfrac{1}{\sin ^2 \theta}Divide both sides of the equation by \sin ^2 \theta
\displaystyle 1 + \cot ^ 2 \theta\displaystyle =\displaystyle \csc^ 2 \thetaMultiplicative inverse property and reciprocal identities

Example 3

Simplify the expressions.

a

\csc \theta \tan \theta \cos \theta

Worked Solution
Create a strategy

Use reciprocal identities and the tangent identity to rewrite the trigonometric functions in terms of sine and cosine.

Apply the idea
\displaystyle \csc \theta \tan \theta \cos \theta\displaystyle =\displaystyle \dfrac{1}{\sin \theta} \cdot \dfrac{\sin \theta}{\cos \theta} \cdot \cos \thetaReciprocal identities and tangent identity
\displaystyle =\displaystyle \dfrac{\cos \theta}{\sin \theta} \cdot \dfrac{\sin \theta}{\cos \theta}Definition of multiplying rational expressions and commutative property of multiplication
\displaystyle =\displaystyle 1Multiplicative inverse property
b

\sec ^2 \theta \left( \cos^2 \theta -1 \right)

Worked Solution
Create a strategy

Since \cos ^2 \theta - 1 is close to one of the Pythagorean identities, we can rewrite the identity to show what \cos ^2 \theta - 1 is equivalent to:

\displaystyle \sin^2 \theta + \cos ^2 \theta\displaystyle =\displaystyle 1Pythagorean identity
\displaystyle \cos ^2 \theta - 1\displaystyle =\displaystyle -\sin^2 \thetaSubtract 1 and \sin^2 \theta from both sides of the equation

Then, use Pythagorean identities and tangent identity to rewrite the trigonometric functions in terms of sine and cosine.

Apply the idea
\displaystyle \sec ^2 \theta \left( \cos^2 \theta -1 \right)\displaystyle =\displaystyle \left( \tan^2 \theta + 1 \right) \left(- \sin^2 \theta \right)Pythagorean identities
\displaystyle =\displaystyle \left( \dfrac{ \sin^2 \theta }{\cos^2 \theta} + 1 \right) \left( -\sin^2 \theta \right)Tangent identity
\displaystyle =\displaystyle \left( \dfrac{ \sin^2 \theta + \cos^2 \theta }{\cos^2 \theta } \right) \left( -\sin^2 \theta \right)Definition of adding rational expressions
\displaystyle =\displaystyle \left( \dfrac{1}{\cos^2 \theta } \right)\left( -\sin^2 \theta \right)Pythagorean identity
\displaystyle =\displaystyle \dfrac{-\sin ^2 \theta }{\cos^2 \theta }Definition of multiplying rational expressions
\displaystyle =\displaystyle -\tan^2 \theta Tangent identity

Example 4

Consider \sin \theta = \dfrac{-2}{3}, where 0 < \theta < 2 \pi.

a

State any quadrant angle \theta lies in.

Worked Solution
Create a strategy

Since \sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}}, and the hypotenuse always represents positive value of length, -2 represents the side length opposite to \theta or the y-coordinate and 3 represents the hypotenuse of the triangle on the coordinate plane. Since -2 is negative, the angle must be in any quadrant where the y-coordinate is negative.

Apply the idea

Quadrants 3 or 4.

b

Find \cos \theta when \pi < \theta < \dfrac{3 \pi}{2}.

Worked Solution
Create a strategy

Note that \theta is restricted to the third quadrant. Use the Pythagorean identity to solve for \cos \theta.

Apply the idea
\displaystyle \sin^{2} \theta + \cos^{2} \theta\displaystyle =\displaystyle 1Pythagorean identity
\displaystyle \left(\dfrac{-2}{3}\right)^{2} + \cos^{2} \theta\displaystyle =\displaystyle 1Substitute \sin \theta = \dfrac{-2}{3}
\displaystyle \dfrac{4}{9} + \cos^{2} \theta\displaystyle =\displaystyle 1Evaluate the square
\displaystyle \cos ^2 \theta\displaystyle =\displaystyle 1 - \dfrac{4}{9}Subtract \dfrac{4}{9} from both sides
\displaystyle \cos^{2} \theta\displaystyle =\displaystyle \dfrac{5}{9}Evaluate the subtraction
\displaystyle \cos \theta\displaystyle =\displaystyle \pm \dfrac{\sqrt{5}}{3}Take the square root of both sides

Since cosine is negative in the third quadrant, \cos \theta= - \dfrac{\sqrt{5}}{3}.

Reflect and check

We could use the Pythagorean theorem to draw a right triangle and find its missing side, then use the domain of the cosine function to determine the sign of the ratio.

A right triangle with legs of length a and b and hypotenuse of length c. The angle opposite the side of length b is labeled theta.
\displaystyle a^2 + b^2\displaystyle =\displaystyle c^2Pythagorean theorem
\displaystyle a^2+2^2\displaystyle =\displaystyle 3^2Substitution
\displaystyle a^2+4\displaystyle =\displaystyle 9Evaluate the exponents
\displaystyle a^2\displaystyle =\displaystyle 5Subtract 4 from both sides of the equation
\displaystyle a\displaystyle =\displaystyle \sqrt{5}Evaluate the square root of both sides of the equation

\cos \theta = \dfrac{\sqrt{5}}{3}, and since cosine is negative on the coordinate plane when 0 < \theta < \dfrac{3 \pi}{2}, \cos \theta= - \dfrac{\sqrt{5}}{3}.

c

Find the value of \tan \theta.

Worked Solution
Create a strategy

Assuming that \theta is in the third quadrant, we can use the tangent identity to determine \tan \theta.

Apply the idea

\tan \theta = \dfrac{\sin \theta}{\cos \theta} = \dfrac{ \dfrac{-2}{3}}{ -\dfrac{\sqrt{5}}{3}} = \dfrac{-2}{3} \cdot - \dfrac{3}{\sqrt{5}}= \dfrac{2}{\sqrt{5}}

Reflect and check

We could also use the side lengths of the right triangle drawn in part (b) to determine \tan \theta in the third quadrant.

\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{2}{\sqrt{5}}. In the third quadrant, \tan \theta is positive, since both sine and cosine are negative and \tan \theta = \dfrac{\sin \theta}{\cos \theta}.

Idea summary

We can write equivalent expressions and find trigonometric function values using a combination of reciprocal identities and the Pythagorean identities given here:\begin{aligned} \sin^2 \theta + \cos^2 \theta = 1 \\\ \tan ^2 \theta + 1 = \sec^2 \theta \\\ 1 + \cot^2 \theta = \csc^2 \theta \end{aligned}

Outcomes

F.TF.C.8

Prove the Pythagorean identity sin^2(θ) cos^2(θ) = 1 and use it to find sin(θ), cos(θ), or tan(θ) given sin(θ), cos(θ), or tan(θ) and the quadrant of the angle.

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