Grade 12 · MHF4U · Advanced Functions

MHF4U: Proving Trig Identities

A trig identity is an equation that holds for every angle where both sides are defined. Proving one means turning one side into the other, one legal step at a time.

How It Works

The rule that makes it a proof: work on each side separately. Write L.S. and R.S., transform one (or both) until they match, and never move terms across the equals sign. Moving terms across assumes the thing you are trying to prove.

Your toolbox, all from the lesson:

Pythagorean: $\sin^2 x + \cos^2 x = 1$ , and dividing it through gives $1 + \tan^2 x = \sec^2 x$ and $1 + \cot^2 x = \csc^2 x$ .

Quotient and reciprocal: $\tan x = \frac{\sin x}{\cos x}$ , $\sec x = \frac{1}{\cos x}$ , $\csc x = \frac{1}{\sin x}$ , $\cot x = \frac{\cos x}{\sin x}$ .

Double angle: $\sin 2x = 2\sin x\cos x$ and $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2\sin^2 x = 2\cos^2 x - 1$ .

The moves that crack most proofs:

Start on the messier side. Short sides are where you land, not where you start.

Rewrite everything in $\sin$ and $\cos$ . It looks longer for a line and then collapses.

Combine fractions over a common denominator, and factor: difference of squares hides everywhere, like $1 - \sin^2 x = (1 - \sin x)(1 + \sin x)$ .

Multiply by a form of 1, like $\frac{1 + \cos x}{1 + \cos x}$ , to make a Pythagorean identity appear.

For $\cos 2x$ , pick the version that cancels with what is already on the page.

To show something is not an identity, one counterexample is enough: a single angle where the two sides differ.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Prove $\tan x\cos x = \sin x$ .
Show solution
1. L.S. $= \tan x\cos x = \frac{\sin x}{\cos x} \cdot \cos x$ .
2. The $\cos x$ cancels: L.S. $= \sin x =$ R.S.
Answer: L.S. = R.S., so $\tan x\cos x = \sin x$ is an identity.
Easy
Prove $\frac{\sin^2 x}{1 - \cos x} = 1 + \cos x$ .
Show solution
1. Start on the left, the messier side. Replace $\sin^2 x$ with $1 - \cos^2 x$ : L.S. $= \frac{1 - \cos^2 x}{1 - \cos x}$ .
2. Factor the difference of squares: $\frac{(1 - \cos x)(1 + \cos x)}{1 - \cos x}$ .
3. Cancel $(1 - \cos x)$ : L.S. $= 1 + \cos x =$ R.S.
Answer: L.S. = R.S., so the identity holds wherever $\cos x \ne 1$ .
Medium
Prove $\frac{1}{1 - \sin x} + \frac{1}{1 + \sin x} = 2\sec^2 x$ .
Show solution
1. Common denominator on the left: L.S. $= \frac{(1 + \sin x) + (1 - \sin x)}{(1 - \sin x)(1 + \sin x)}$ .
2. Simplify the top and expand the bottom: $\frac{2}{1 - \sin^2 x}$ .
3. Pythagorean identity: $1 - \sin^2 x = \cos^2 x$ , so L.S. $= \frac{2}{\cos^2 x}$ .
4. Reciprocal identity: $\frac{2}{\cos^2 x} = 2\sec^2 x =$ R.S.
Answer: L.S. = R.S., so the identity holds.
Medium
Prove $\cos^4 x - \sin^4 x = \cos 2x$ .
Show solution
1. The left side is a difference of squares: L.S. $= (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x)$ .
2. The second bracket is $1$ : L.S. $= \cos^2 x - \sin^2 x$ .
3. That is one form of the double angle identity: L.S. $= \cos 2x =$ R.S.
Answer: L.S. = R.S., so the identity holds.
Test level
Prove $\frac{\sin 2x}{\sin x} - \frac{\cos 2x}{\cos x} = \sec x$ .
Show solution
1. First term: $\frac{\sin 2x}{\sin x} = \frac{2\sin x\cos x}{\sin x} = 2\cos x$ .
2. Second term: the denominator is $\cos x$ , so choose $\cos 2x = 2\cos^2 x - 1$ . L.S. $= 2\cos x - \frac{2\cos^2 x - 1}{\cos x}$ .
3. Common denominator: L.S. $= \frac{2\cos^2 x - (2\cos^2 x - 1)}{\cos x} = \frac{1}{\cos x}$ .
4. Reciprocal identity: L.S. $= \sec x =$ R.S.
Answer: L.S. = R.S., so the identity holds wherever $\sin x \ne 0$ and $\cos x \ne 0$ .

Learn It on Martian Lab

Martian Lab teaches this in the Trigonometric Identities lesson of its MHF4U Trigonometric Ratios unit. Watch it, then take the Trigonometric Ratios practice tests to see if it stuck.

Watch the Trigonometric Identities lesson free

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Free Trigonometric Ratios practice test, with solutions

More MHF4U topics: Long and Synthetic Division · Vertical and Horizontal Asymptotes · Solving Trig Equations · Laws of Logarithms · Solving Logarithmic Equations

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