Pure 5.8 - Proofs with trigonometric identities | Edexcel A-Level Mathematics

Prove trigonometric identities by recognising compound-angle patterns, choosing useful algebra and handling reciprocal fractions. You will keep a valid chain of reasoning, explain necessary domain restrictions, and distinguish a proof from a numerical check.

What a trigonometric identity proof means

If a trig statement works at several angles, have you proved it? Consider $\sin x+\cos x=1$ . It works at $0$ , $\pi/2$ and $2\pi$ . Yet at $x=\pi/4$ , the left side is $\sqrt2$ , not $1$ . Several successful checks can hide a false claim.

Here is a useful surprise: that false statement even works at infinitely many angles, because its successful inputs repeat every $2\pi$ . Infinitely many examples still do not cover every real angle.

You already know

A counterexample disproves a universal claim. Algebra can prove a universal claim when every transformation is valid for every input in the stated domain.

An identity is about expressions representing the same value throughout their common domain, rather than about finding selected roots.

Trigonometric identity

An identity is an equality true for every input for which the expressions being compared are defined. Write $\equiv$ to emphasise this claim. For example, $\sin^2x+\cos^2x\equiv1$ for every real $x$ . The statement $\sin x=1/2$ is an equation with particular solutions, not an identity.

A practical proof starts with one side and changes its form until it becomes the other. Each equals sign must connect expressions that really have equal values on the domain.

Key idea

Numerical substitution can catch an error, but an identity proof needs reasoning that covers the whole stated domain.

Decide whether the evidence really supports an identity.

Choose a route before writing

The target expression is a clue. Products such as $\cos A\cos B+\sin A\sin B$ suggest a compound angle; reciprocal functions suggest a common denominator. Begin with the side whose structure you can simplify.

Core identities for proof

\tan x=\frac{\sin x}{\cos x}, \quad\cot x=\frac{\cos x}{\sin x}, \quad \sec x=\frac1{\cos x}, \quad\operatorname{cosec}x=\frac1{\sin x}

where the relevant denominators are non-zero. Also,

\sin^2x+\cos^2x=1, \quad \sec^2x=1+\tan^2x, \quad \operatorname{cosec}^2x=1+\cot^2x

on their respective domains, and $\sin2x=2\sin x\cos x$ for every real $x$ .

Not in the formula booklet: learn it.

By contrast, $\cos(A-B)=\cos A\cos B+\sin A\sin B$ is in the formula booklet. Recognition can make a long-looking proof very short.

Worked example 1: read a product pattern backwards

Question. Prove $\cos x\cos2x+\sin x\sin2x\equiv\cos x$ .

Plan. Recognise cosine difference with $A=2x$ , $B=x$ .

\begin{aligned} \cos x\cos2x+\sin x\sin2x &=\cos2x\cos x+\sin2x\sin x && \text{reorder products}\\ &=\cos(2x-x) && \text{cosine difference}\\ &=\cos x && \text{simplify the input} \end{aligned}

Why use the compound formula? Its products match exactly, avoiding unnecessary expansions.

Answer. The identity is proved for all real $x$ .

Check. At $x=\pi/3$ , the left side is $-1/4+3/4=1/2$ , matching $\cos x$ .

Change only the multiple from $2x$ to $3x$ . The same recognition still works.

Worked example 2: the changed multiple

Question. Prove $\cos x\cos3x+\sin x\sin3x\equiv\cos2x$ .

Plan. Combine the two products as cosine difference.

\begin{aligned} \cos x\cos3x+\sin x\sin3x &=\cos(3x-x) && \text{use the supplied identity}\\ &=\cos2x && \text{subtract the inputs} \end{aligned}

Why choose $3x-x$ ? The two original input angles determine the difference; do not multiply the angles together.

Answer. The identity holds for all real $x$ .

Check. At $x=\pi/3$ , the left side is $-1/2$ , matching $\cos(2\pi/3)$ .

When reciprocals replace the product pattern, a different route is cleaner. The following pair shows how to create the double angle appearing in the target.

Worked example 3: make a common denominator

Question. Prove $\tan\theta+\cot\theta\equiv2\operatorname{cosec}2\theta$ , where both sides are defined.

Plan. Convert to sine and cosine, combine, then recognise $\sin2\theta$ .

\begin{aligned} \tan\theta+\cot\theta &=\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta} && \text{use quotient forms}\\ &=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta} && \text{combine fractions}\\ &=\frac1{\sin\theta\cos\theta} && \text{Pythagorean identity}\\ &=\frac2{\sin2\theta}=2\operatorname{cosec}2\theta && \text{double angle} \end{aligned}

Why create a product in the denominator? The target contains the reciprocal of $\sin2\theta$ , which equals $2\sin\theta\cos\theta$ .

Answer. The identity is proved when $\sin\theta\cos\theta\ne0$ .

Check. At $\theta=\pi/4$ , both sides equal $2$ .

Change only addition to subtraction, keeping the same first common-denominator step.

Worked example 4: a difference of quotients

Question. Prove $\tan\theta-\cot\theta\equiv-2\cot2\theta$ , on the common domain.

Plan. Combine the fractions, then recognise cosine and sine double angles.

\begin{aligned} \tan\theta-\cot\theta &=\frac{\sin^2\theta-\cos^2\theta}{\sin\theta\cos\theta} && \text{combine quotients}\\ &=\frac{-\cos2\theta}{\frac12\sin2\theta} && \text{double-angle forms}\\ &=-2\cot2\theta && \text{use the quotient definition} \end{aligned}

Why the minus sign? Cosine double angle is $\cos^2\theta-\sin^2\theta$ , the reverse of this numerator.

Answer. The identity holds when $\sin\theta\cos\theta\ne0$ .

Check. At $\theta=\pi/4$ , both sides are zero.

Choose the route by reading the expression, not by applying every identity you remember.

Problem-solving

Fractions suggest one denominator; squares suggest Pythagoras; products of two different angles suggest compound angles. Let the target tell you which form would help.

Use the fraction route with squared reciprocals.

Create the target with algebra

Sometimes the missing ingredient is ordinary algebra. A denominator $1+\sin\theta$ pairs naturally with $1-\sin\theta$ , since their product is $1-\sin^2\theta=\cos^2\theta$ .

Multiplying by one

If $g(\theta)\ne0$ , then multiplying numerator and denominator by $g(\theta)$ leaves a fraction's value unchanged. The factors $1+\sin\theta$ and $1-\sin\theta$ are a useful difference-of-squares pair. Multiplying only the numerator would change the value.

Check that the new factor is non-zero on the identity's common domain before introducing it.

Worked example 5: create the target numerator

Question. Prove $\frac{\cos\theta}{1+\sin\theta}\equiv\frac{1-\sin\theta}{\cos\theta}$ , where both sides are defined.

Plan. Multiply the left fraction by $(1-\sin\theta)/(1-\sin\theta)$ .

On the common domain, $\cos\theta\ne0$ , hence $\sin\theta\ne\pm1$ . The new divisor is non-zero.

\begin{aligned} \frac{\cos\theta}{1+\sin\theta} &=\frac{\cos\theta(1-\sin\theta)}{(1+\sin\theta)(1-\sin\theta)} && \text{multiply by one}\\ &=\frac{\cos\theta(1-\sin\theta)}{1-\sin^2\theta} && \text{difference of squares}\\ &=\frac{\cos\theta(1-\sin\theta)}{\cos^2\theta} && \text{Pythagoras}\\ &=\frac{1-\sin\theta}{\cos\theta} && \text{cancel a non-zero factor} \end{aligned}

Why choose the opposite sign? It creates a difference of squares and the exact target numerator.

Answer. The identity is proved on the common domain $\cos\theta\ne0$ .

Check. At $\theta=\pi/6$ , both sides equal $1/\sqrt3$ .

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Pure Mathematics (Papers 1 and 2)

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Mechanics (Paper 3, Section B)

Pure 5.8 - Proofs with trigonometric identities

What a trigonometric identity proof means

Choose a route before writing

Create the target with algebra

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