Math Core

Some equations look like they could be identities, and some really are. To verify an identity means to prove that the two sides are equal for every allowed value of the variable. It is a lot like a geometry proof: you start from something you know and reach the goal one justified step at a time.

The rules of the game

When you verify an identity, you are not solving an equation. You don't know yet that the two sides are equal, so you can't treat the statement as a true equation.

How to verify an identity

Pick one side (usually the more complicated one) and transform it, step by step, using known identities and algebra, until it looks exactly like the other side. Don't move terms across the equals sign, and don't do the same thing to both sides.

Why not work on both sides at once? Because operations like squaring both sides can turn a false statement into a true one. For example, $-1 = 1$ is false, but squaring both sides gives $1 = 1$ . Working on one side avoids that trap. Each step replaces an expression with one that is truly equal to it, so the chain proves the two ends are equal.

A toolbox of strategies

There is no single recipe, but these moves work most of the time. Try them roughly in this order.

Start with the more complicated side. It is easier to simplify than to "complicate."
Rewrite in sines and cosines if you see tangent, secant, cosecant or cotangent mixed together.
Look for Pythagorean patterns such as $1 - \sin^2 x$ , $\sec^2 x - 1$ or $1 + \cot^2 x$ .
Combine fractions over a common denominator, or split one fraction into two.
Factor: common factors, differences of squares, even trinomials in $\sin x$ .
Multiply by a conjugate such as $\dfrac{1 + \sin x}{1 + \sin x}$ when a denominator is $1 - \sin x$ . The product $(1 - \sin x)(1 + \sin x) = \cos^2 x$ often unlocks the problem.
Keep an eye on the goal. If the other side has only $\sin x$ in it, steer toward sines.

Writing it up

A clean way to present the work is a vertical chain. Each line is equal to the one before it, and you can note the reason on the right, like the two columns of a geometry proof.

Worked example: Rewrite in sines and cosines

Verify: $\sec x - \cos x = \sin x \tan x$ .

The left side has a secant, so rewrite it and combine the fractions.

\begin{aligned} \sec x - \cos x &= \frac{1}{\cos x} - \cos x && \text{reciprocal identity} \\ &= \frac{1}{\cos x} - \frac{\cos^2 x}{\cos x} && \text{common denominator} \\ &= \frac{1 - \cos^2 x}{\cos x} && \text{subtract} \\ &= \frac{\sin^2 x}{\cos x} && \text{Pythagorean identity} \\ &= \sin x \cdot \frac{\sin x}{\cos x} && \text{split the product} \\ &= \sin x \tan x && \text{quotient identity} \end{aligned}

The left side became the right side, so the identity is verified.

Worked example: Combine fractions

Verify: $\dfrac{\sin x}{1 + \cos x} + \dfrac{1 + \cos x}{\sin x} = 2\csc x$ .

Work on the left side. The common denominator is $\sin x\,(1 + \cos x)$ .

\begin{aligned} \frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x} &= \frac{\sin^2 x + (1 + \cos x)^2}{\sin x\,(1 + \cos x)} && \text{common denominator} \\ &= \frac{\sin^2 x + 1 + 2\cos x + \cos^2 x}{\sin x\,(1 + \cos x)} && \text{expand the square} \\ &= \frac{2 + 2\cos x}{\sin x\,(1 + \cos x)} && \sin^2 x + \cos^2 x = 1 \\ &= \frac{2(1 + \cos x)}{\sin x\,(1 + \cos x)} && \text{factor} \\ &= \frac{2}{\sin x} = 2\csc x && \text{cancel} \end{aligned}

Worked example: Multiply by a conjugate

Verify: $\dfrac{\cos x}{1 - \sin x} = \sec x + \tan x$ .

The left side has $1 - \sin x$ in the denominator. Multiply the top and bottom by its conjugate, $1 + \sin x$ .

\begin{aligned} \frac{\cos x}{1 - \sin x} &= \frac{\cos x\,(1 + \sin x)}{(1 - \sin x)(1 + \sin x)} && \text{multiply by } \tfrac{1 + \sin x}{1 + \sin x} \\ &= \frac{\cos x\,(1 + \sin x)}{1 - \sin^2 x} && \text{difference of squares} \\ &= \frac{\cos x\,(1 + \sin x)}{\cos^2 x} && \text{Pythagorean identity} \\ &= \frac{1 + \sin x}{\cos x} && \text{cancel one } \cos x \\ &= \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \sec x + \tan x && \text{split the fraction} \end{aligned}

Common mistake

Don't "cross multiply" or add the same thing to both sides, as if you were solving an equation. That assumes the identity is already true, which is exactly what you are trying to prove. Transform one side until it matches the other.

Is it really an identity?

Before you spend ten minutes on a proof, test the claim with a number. If the two sides give different values for even one allowed angle, the equation is not an identity.

Worked example: Testing with a value

Is $\sin x + \cos x = 1$ an identity?

Try $x = 45^\circ$ : the left side is $\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2} = \sqrt{2} \approx 1.414$ , which is not $1$ . So it is not an identity. (It is true for some angles, such as $x = 0^\circ$ , but that only makes it an equation.)

Tip

A numerical check can show that something is not an identity, but it can never prove that something is one. Matching values at a few angles is encouraging; only an algebraic chain is a proof.

Practice

Practice 1

You want to verify $\dfrac{\tan x}{\sec x} = \sin x$ . Which is the best first step?

Practice 2

To verify $\dfrac{1}{1 + \cos x} = \dfrac{1 - \cos x}{\sin^2 x}$ , a student multiplies the left side by $\dfrac{1 - \cos x}{1 - \cos x}$ and gets $\dfrac{1 - \cos x}{1 - \cos^2 x}$ . Which step comes next?

Practice 3

The expression $\cot^2 x\,(\sec^2 x - 1)$ equals a constant wherever it is defined. What is that constant?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Which equation is not an identity?

Practice 5

The expression $\dfrac{1}{1 - \sin x} + \dfrac{1}{1 + \sin x}$ equals $k\sec^2 x$ for some constant $k$ . What is $k$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Which expression is equal to $\dfrac{\sin^4 x - \cos^4 x}{\sin^2 x - \cos^2 x}$ (where the denominator is not $0$ )?

Practice 7

To verify $\dfrac{1 - \sin x}{\cos x} = \dfrac{\cos x}{1 + \sin x}$ , which strategy leads most directly to the right side?