Topics
10. Trigonometry
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United States of America
Grades 9-12

10.05 Trigonometric identities

Lesson
Practice

Introduction

We will use the Pythagorean theorem and trigonometric functions to manipulate trigonometric 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 identity can also be used to find trigonometric function values given certain information: \left(\sin \theta \right)^2 + \left(\cos \theta \right) ^2 = 1 \quad \text{ or } \quad \sin^2 \theta + \cos^2 \theta = 1

Recall the basic trigonometric identities:

\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}

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

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 identity given here: \sin^2 \theta + \cos^2 \theta = 1

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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