Verifying Trigonometric Identities

Quick answer

An identity is an equation that holds for every value where both sides are defined, and verifying one means turning a single side into the other by steps that are equalities at every stage. The working tools are the reciprocal, quotient and Pythagorean identities, along with common denominators, factoring and conjugates. Operating on both sides assumes what is being proved, so the work stays on one side. One counterexample is enough to show that a claim is not an identity.

What you'll learn

State the reciprocal, quotient and Pythagorean identities
Verify an identity by transforming one side into the other
Choose a strategy: common denominators, factoring or a conjugate
Disprove a false claim with a single counterexample

What an identity claims

The equation $\sin x = \tfrac{1}{2}$ is true for some angles and false for others; solving it means finding which. The equation $\sin^2 x + \cos^2 x = 1$ is different. It is true for every angle, and there is nothing to solve. An equation like that is an identity, and the work is to prove it.

Three families of identities do most of the work.

Family	Identities
Reciprocal	$\csc x = \tfrac{1}{\sin x}$ , $\sec x = \tfrac{1}{\cos x}$ , $\cot x = \tfrac{1}{\tan x}$
Quotient	$\tan x = \tfrac{\sin x}{\cos x}$ , $\cot x = \tfrac{\cos x}{\sin x}$
Pythagorean	$\sin^2 x + \cos^2 x = 1$ , $1 + \tan^2 x = \sec^2 x$ , $1 + \cot^2 x = \csc^2 x$

The second and third Pythagorean identities come from dividing the first by $\cos^2 x$ and by $\sin^2 x$ .

The rules of the game

Verifying an identity means starting on one side and rewriting it, one equality at a time, until it reads as the other side. Working on both sides at once is not allowed, since that assumes the two sides are equal.

Six moves cover almost every problem:

Begin with the more complicated side.
Rewrite everything in sines and cosines.
Combine fractions over a common denominator.
Factor, and look for a difference of squares.
Multiply a fraction above and below by a conjugate.
Trade $\sin^2 x$ for $1 - \cos^2 x$ , or the other way round.

Why one-sided work is the honest proof

Suppose the claim were $\sin x = -\sin x$ , which is false at $x = \tfrac{\pi}{2}$ . Squaring both sides gives $\sin^2 x = \sin^2 x$ , a true statement. So arriving at something true proves nothing about where you started: a false claim can lead to a true one. Transforming a single side avoids that trap, because no step uses the claim. Each step replaces an expression by an equal expression, so the finished chain of equalities proves the two sides agree wherever both are defined.

Worked examples

Common mistakes

Practice problems

Verify $\sin x \sec x = \tan x$ .

Answer

$\sin x \cdot \tfrac{1}{\cos x} = \tfrac{\sin x}{\cos x} = \tan x$

Full solution

Rewrite $\sec x$ as $\tfrac{1}{\cos x}$ . The product is $\tfrac{\sin x}{\cos x}$ , which is the quotient identity for $\tan x$ .
Verify $\left(1 + \tan^2 x\right)\cos^2 x = 1$ .

Answer

$\sec^2 x \cos^2 x = 1$

Full solution

The Pythagorean identity gives $1 + \tan^2 x = \sec^2 x$ , and $\sec^2 x \cos^2 x = \tfrac{\cos^2 x}{\cos^2 x} = 1$ .
Verify $\cos x \csc x = \cot x$ .

Answer

$\cos x \cdot \tfrac{1}{\sin x} = \cot x$

Full solution

Write $\csc x$ as $\tfrac{1}{\sin x}$ . The product is $\tfrac{\cos x}{\sin x}$ , which is $\cot x$ .
Verify $(1 - \cos x)(1 + \cos x) = \sin^2 x$ .

Answer

$1 - \cos^2 x = \sin^2 x$

Full solution

The left side is a difference of squares, $1 - \cos^2 x$ , and the Pythagorean identity rewrites it as $\sin^2 x$ .
Verify $\sec^4 x - \tan^4 x = \sec^2 x + \tan^2 x$ .

Hint

Factor the left side as a difference of squares.

Answer

$\left(\sec^2 x - \tan^2 x\right)\left(\sec^2 x + \tan^2 x\right) = \sec^2 x + \tan^2 x$

Full solution

Factoring gives two factors, and $\sec^2 x - \tan^2 x = 1$ by the Pythagorean identity, so the product is the remaining factor.
Verify $(\sin x + \cos x)^2 = 1 + 2\sin x \cos x$ .

Answer

$\sin^2 x + 2\sin x\cos x + \cos^2 x = 1 + 2\sin x\cos x$

Full solution

Expand the square, then replace $\sin^2 x + \cos^2 x$ with $1$ .
Verify $\tan x + \cot x = \sec x \csc x$ .

Answer

$\tfrac{\sin x}{\cos x} + \tfrac{\cos x}{\sin x} = \tfrac{1}{\sin x \cos x}$

Full solution

Over the common denominator $\sin x \cos x$ , the sum is $\tfrac{\sin^2 x + \cos^2 x}{\sin x\cos x} = \tfrac{1}{\sin x\cos x}$ , which is $\sec x \csc x$ .
Verify $\displaystyle\frac{\cos x}{1 + \sin x} = \frac{1 - \sin x}{\cos x}$ .

Hint

Multiply above and below by $1 - \sin x$ .

Answer

$\tfrac{\cos x(1 - \sin x)}{1 - \sin^2 x} = \tfrac{1 - \sin x}{\cos x}$

Full solution

The conjugate turns the denominator into $1 - \sin^2 x = \cos^2 x$ . One factor of $\cos x$ cancels, leaving $\tfrac{1 - \sin x}{\cos x}$ .
Show that $\sin x = \cos x$ is not an identity.

Answer

At $x = 0$ : $\sin 0 = 0$ and $\cos 0 = 1$ .

Full solution

One value where the sides differ is enough. The two are equal only at angles such as $\tfrac{\pi}{4}$ , where both are $\tfrac{\sqrt{2}}{2}$ , so the equation is one to solve, not an identity to prove.
A student verifies $\sin x = -\sin x$ by squaring both sides to get $\sin^2 x = \sin^2 x$ , and concludes it is an identity. What went wrong?

Hint

Test $x = \tfrac{\pi}{2}$ .

Answer

Squaring is not reversible, and the claim is false: at $x = \tfrac{\pi}{2}$ the sides are $1$ and $-1$ .

Full solution

Operating on both sides assumes the claim. Squaring loses the sign, so both sides become the same expression whether or not the original claim holds. A one-sided rewrite could never produce this, since it never uses the claim.

Frequently asked questions

What is the difference between an identity and an equation?

An equation asks which values make it true, and usually only some do. An identity is true for every value where both sides are defined.

How do you verify a trigonometric identity?

Pick one side, usually the more complicated one, and rewrite it step by step until it matches the other side. Every step must be an equality.

Why can't I work on both sides at once?

Operating on both sides assumes the two sides are already equal, which is what you are trying to prove. Squaring both sides of the false claim sin x = −sin x gives a true statement.

What are the three Pythagorean identities?

sin²x + cos²x = 1, 1 + tan²x = sec²x, and 1 + cot²x = csc²x. The second and third come from dividing the first by cos²x and by sin²x.

How do I show something is not an identity?

Find one value that breaks it. sin x + cos x = 1 fails at x = π/4, where the left side is about 1.414.

What to learn next

Sum and Difference Formulas for Sine, Cosine and TangentWhy cos(α − β) = cos α cos β + sin α sin β, proved by rotating a chord of the unit circle, how the other formulas follow, and how to find exact values like cos 15°.

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

CCSS.MATH.CONTENT.HSF.TF.C.8Trigonometric FunctionsProve the Pythagorean identity sin²(θ) + cos²(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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