AP®MathsCollege BoardPrecalculusRevision NotesTrigonometric & Polar FunctionsTrigonometric Equations, Inequalities & IdentitiesVerifying Additional Trigonometric Identities

Verifying Additional Trigonometric Identities (College Board AP® Precalculus): Revision Note

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 15 April 2026

Verifying additional trigonometric identities

What does it mean to verify a trigonometric identity?

A trigonometric identity is
- an equation involving trigonometric functions
- that is true for all values of the variable in the common domain of both sides
To verify an identity means to show that the two sides of the equation are equivalent
- i.e. that one side can be rewritten to match the other
- using known identities and algebraic manipulations
Verifying an identity is not the same as solving an equation
- You do not move terms from one side to the other
- Instead, you transform one side step by step until it looks exactly like the other side

What tools are used to verify identities?

Any identity or algebraic rule already known can be used, including
- The Pythagorean identity and its rearranged forms
  - $\sin^{2} θ + \cos^{2} θ = 1$ , $\sec^{2} θ - \tan^{2} θ = 1$ , $\csc^{2} θ - \cot^{2} θ = 1$
- The sum, difference, and double-angle identities
- The reciprocal relationships
  - $\sec θ = \frac{1}{\cos θ}$ , $\csc θ = \frac{1}{\sin θ}$ , $\cot θ = \frac{1}{\tan θ}$
- The quotient identities
  - $\tan θ = \frac{\sin θ}{\cos θ}$ , $\cot θ = \frac{\cos θ}{\sin θ}$
- Standard algebraic manipulations
  - combining fractions, factoring, expanding, distributing, etc.

What is the general process for verifying an identity?

Start by picking one side to work with
- It is usually best to start with the more complex side
  - since simplifying is typically easier than complicating
Look at the target side
- I.e. the side you'll need to transform your 'starting side' into
- What form does it take?
  - This often suggests which identities or manipulations will be useful
- E.g. if the target side contains only $\sin$ and $\cos$
  - consider converting all the functions on the starting side into $\sin$ and $\cos$
- Or if the target side contains a single trig function
  - look for a way to cancel or combine terms on the starting side
Transform the starting side step by step, using known identities and algebraic rules
- until it matches the target side exactly
At each step, write clearly which identity or rule is being applied

What strategies are commonly useful?

Convert to sine and cosine
- Rewriting all trig functions on one side in terms of $\sin$ and $\cos$ is often a good way to start
  - particularly when the identity involves reciprocal functions
Combine fractions
- If one side has multiple fractions, combining them over a common denominator often reveals opportunities for further simplification
Look for Pythagorean patterns
- Expressions like $1 - \sin^{2} x$ , $1 + \tan^{2} x$ , or $(1 - \cos x) (1 + \cos x)$ can often be simplified using the Pythagorean identity
Factor when possible
- Factoring numerators or denominators can reveal cancellations
Work from both sides if necessary
- If it is not clear how to proceed from one side, you can work on both sides separately to get them into a common simpler form
- But the final write-up should still show one side being transformed into the other

Examiner Tips and Tricks

When verifying an identity, do not move terms across the equals sign as you would when solving an equation.

An identity verification is a proof of equivalence
Treating it as an equation to solve is one of the most common mistakes

Worked Example

Verify the identity:

$\frac{\sin x}{1 - \cos x} + \frac{\sin x}{1 + \cos x} = 2 \csc x$

Answer

Start with the left-hand side and simplify step by step

First combine the two fractions over a common denominator

$\frac{\sin x}{1 - \cos x} + \frac{\sin x}{1 + \cos x} = \frac{\sin x (1 + \cos x) + \sin x (1 - \cos x)}{(1 - \cos x) (1 + \cos x)}$

Expand the brackets in the numerator and denominator

Note that the denominator is a 'difference of two squares'

$= \frac{\sin x + \sin x \cos x + \sin x - \sin x \cos x}{1 + \cos x - \cos x - \cos^{2} x} = \frac{2 \sin x}{1 - \cos^{2} x}$

Use the Pythagorean identity in the denominator

i.e. $1 - \cos^{2} x = \sin^{2} x$

$= \frac{2 \sin x}{\sin^{2} x}$

Cancel one factor of $\sin x$

$= \frac{2}{\sin x}$

Convert to the reciprocal function to match the target

i.e. using $\csc x = \frac{1}{\sin x}$

$= 2 \csc x$

The left hand side has been transformed step by step into the right-hand side, so the identity is verified

$\frac{\sin x}{1 - \cos x} + \frac{\sin x}{1 + \cos x} = 2 \csc x$

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