Proving Trigonometric Identities

Builds on: aqa-alevel-maths-pure-1 / trigonometry — this lesson is the A2-level deep dive.

This lesson focuses on strategies and techniques for proving trigonometric identities. Identity proofs appear frequently in A-Level examinations, and mastering them requires both a solid knowledge of standard identities and systematic problem-solving strategies.

Spec Mapping — AQA 7357 Section E Trigonometry. This lesson covers the proof techniques for trigonometric identities using Pythagorean, compound and double-angle results content of Section E, at the depth required for A2-level synoptic questions. Refer to the official AQA specification document for exact wording.

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What Is a Trigonometric Identity?

A trigonometric identity is an equation that is true for all values of the variable in its domain. We write identities using the symbol ≡ (identically equal to), though the = sign is also accepted in examinations.

For example, sin²θ + cos²θ ≡ 1 is true for every real number θ.

An identity is different from an equation, which is only true for specific values. For example, sin θ = 1/2 is only true for certain values of θ.

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The Toolkit: Identities You Must Know

Fundamental Identities

$$\begin{aligned} \sin^{2}\theta + \cos^{2}\theta &\equiv 1 \\ \tan\theta &\equiv \frac{\sin\theta}{\cos\theta} \end{aligned}$$sin2θ+cos2θtanθ​≡1≡cosθsinθ​​

Pythagorean Identities

$$\begin{aligned} 1 + \tan^{2}\theta &\equiv \sec^{2}\theta \\ 1 + \cot^{2}\theta &\equiv \operatorname{cosec}^{2}\theta \end{aligned}$$1+tan2θ1+cot2θ​≡sec2θ≡cosec2θ​

Double Angle Formulae

$$\begin{aligned} \sin 2A ≡ 2\sin A \cos A \\ \cos 2A ≡ \cos^{2}A - \sin^{2}A ≡ 2\cos^{2}A - 1 ≡ 1 - 2\sin^{2}A \\ \tan 2A ≡ 2\tan A / (1 - \tan^{2}A) \end{aligned}$$sin2A≡2sinAcosAcos2A≡cos2A−sin2A≡2cos2A−1≡1−2sin2Atan2A≡2tanA/(1−tan2A)​

Addition Formulae

$$\begin{aligned} \sin(A \pm B) ≡ \sin A \cos B \pm \cos A \sin B \\ \cos(A \pm B) ≡ \cos A \cos B ∓ \sin A \sin B \\ \tan(A \pm B) ≡ (\tan A \pm \tan B) / (1 ∓ \tan A \tan B) \end{aligned}$$sin(A±B)≡sinAcosB±cosAsinBcos(A±B)≡cosAcosB∓sinAsinBtan(A±B)≡(tanA±tanB)/(1∓tanAtanB)​

Reciprocal Definitions

$$\begin{aligned} \sec\theta &\equiv \frac{1}{\cos\theta} \\ \operatorname{cosec}\theta &\equiv \frac{1}{\sin\theta} \\ \cot\theta &\equiv \frac{\cos\theta}{\sin\theta} \equiv \frac{1}{\tan\theta} \end{aligned}$$secθcosecθcotθ​≡cosθ1​≡sinθ1​≡sinθcosθ​≡tanθ1​​

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General Strategies for Proving Identities

Strategy 1: Work from One Side Only

Choose the more complex side (usually the left-hand side) and manipulate it until it equals the other side. Never move terms from one side to the other — this assumes the identity is true, which is what you are trying to prove.

Strategy 2: Convert Everything to sin and cos

When reciprocal or quotient functions (sec, cosec, cot, tan) appear, replacing them with sin and cos expressions often simplifies the work.

Strategy 3: Look for Pythagorean Identities

If you see sin²θ + cos²θ, replace it with 1. If you see 1 − sin²θ, replace it with cos²θ. If you see sec²θ − 1, replace it with tan²θ. And so on.

Strategy 4: Factorise

Look for common factors, difference of two squares, or expressions that factorise as quadratics.

Strategy 5: Multiply by a Conjugate

If you have a fraction with a binomial denominator like (1 − sin θ), multiply top and bottom by the conjugate (1 + sin θ) to create a difference of squares.

Strategy 6: Combine Fractions

If the expression contains two or more fractions, combine them over a common denominator.

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Specimen Question (Modelled on the AQA Paper Format)

Note: this question is constructed to model AQA Paper 1/2/3 style; it is not a reproduction of any published past paper.

Worked Examples

Example 1: Prove that (1 − sin²θ)(1 + tan²θ) ≡ 1.

$$\begin{aligned} \text{LHS} &= (1 - \sin^{2}\theta)(1 + \tan^{2}\theta) \\ &= \cos^{2}\theta \times \sec^{2}\theta && [\text{using } 1 - \sin^{2}\theta = \cos^{2}\theta \text{ and } 1 + \tan^{2}\theta = \sec^{2}\theta] \\ &= \cos^{2}\theta \times \frac{1}{\cos^{2}\theta} \\ &= 1 \\ &= \text{RHS} \qquad\blacksquare \end{aligned}$$LHS​=(1−sin2θ)(1+tan2θ)=cos2θ×sec2θ=cos2θ×cos2θ1​=1=RHS■​[using 1−sin2θ=cos2θ and 1+tan2θ=sec2θ]

Example 2: Prove that cosec θ − sin θ ≡ cos θ cot θ.

$$\begin{aligned} \text{LHS} &= \frac{1}{\sin\theta} - \sin\theta \\ &= \frac{1 - \sin^{2}\theta}{\sin\theta} \\ &= \frac{\cos^{2}\theta}{\sin\theta} \\ &= \cos\theta \times \frac{\cos\theta}{\sin\theta} \\ &= \cos\theta\cot\theta \\ &= \text{RHS} \quad \blacksquare \end{aligned}$$LHS​=sinθ1​−sinθ=sinθ1−sin2θ​=sinθcos2θ​=cosθ×sinθcosθ​=cosθcotθ=RHS■​

Example 3: Prove that (sec θ − 1)(sec θ + 1) ≡ tan²θ.

$$\begin{aligned} \text{LHS} &= \sec^{2}\theta - 1 && [\text{difference of two squares}] \\ &= \tan^{2}\theta && [\text{using } 1 + \tan^{2}\theta = \sec^{2}\theta, \text{ so } \sec^{2}\theta - 1 = \tan^{2}\theta] \\ &= \text{RHS} \qquad\blacksquare \end{aligned}$$LHS​=sec2θ−1=tan2θ=RHS■​​[difference of two squares][using 1+tan2θ=sec2θ, so sec2θ−1=tan2θ]​

Example 4: Prove that (sin A + cos A)² ≡ 1 + sin 2A.

$$\begin{aligned} \text{LHS} &= \sin^{2}A + 2\sin A \cos A + \cos^{2}A \\ &= (\sin^{2}A + \cos^{2}A) + 2\sin A \cos A \\ &= 1 + \sin 2A \\ &= \text{RHS} \qquad\blacksquare \end{aligned}$$LHS​=sin2A+2sinAcosA+cos2A=(sin2A+cos2A)+2sinAcosA=1+sin2A=RHS■​

Example 5: Prove that tan θ + cot θ ≡ sec θ cosec θ.

$$\begin{aligned} \text{LHS} &= \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} \\ &= \frac{\sin^{2}\theta + \cos^{2}\theta}{\sin\theta \cos\theta} \\ &= \frac{1}{\sin\theta \cos\theta} \\ &= \frac{1}{\sin\theta}\cdot\frac{1}{\cos\theta} \\ &= \operatorname{cosec}\theta \times \sec\theta \\ &= \sec\theta \operatorname{cosec}\theta \\ &= \text{RHS} \qquad\blacksquare \end{aligned}$$LHS​=cosθsinθ​+sinθcosθ​=sinθcosθsin2θ+cos2θ​=sinθcosθ1​=sinθ1​⋅cosθ1​=cosecθ×secθ=secθcosecθ=RHS■​

Example 6: Prove that (1 + cos 2θ)/(sin 2θ) ≡ cot θ.

$$\begin{aligned} \text{LHS} &= \frac{1 + 2\cos^{2}\theta - 1}{2\sin\theta \cos\theta} && [\text{using } \cos 2\theta = 2\cos^{2}\theta - 1] \\ &= \frac{2\cos^{2}\theta}{2\sin\theta \cos\theta} \\ &= \frac{\cos\theta}{\sin\theta} \\ &= \cot\theta \\ &= \text{RHS} \qquad\blacksquare \end{aligned}$$LHS​=2sinθcosθ1+2cos2θ−1​=2sinθcosθ2cos2θ​=sinθcosθ​=cotθ=RHS■​[using cos2θ=2cos2θ−1]

Example 7: Prove that 1/(1 − sin θ) + 1/(1 + sin θ) ≡ 2sec²θ.

$$\begin{aligned} \text{LHS} &= \frac{(1 + \sin\theta) + (1 - \sin\theta)}{(1 - \sin\theta)(1 + \sin\theta)} \\ &= \frac{2}{1 - \sin^{2}\theta} \\ &= \frac{2}{\cos^{2}\theta} \\ &= 2\sec^{2}\theta \\ &= \text{RHS} \qquad\blacksquare \end{aligned}$$LHS​=(1−sinθ)(1+sinθ)(1+sinθ)+(1−sinθ)​=1−sin2θ2​=cos2θ2​=2sec2θ=RHS■​

Example 8: Prove that (cos 2A)/(1 + sin 2A) ≡ (cos A − sin A)/(cos A + sin A).

$$\begin{aligned} \text{LHS} &= \frac{\cos^{2}A - \sin^{2}A}{1 + 2\sin A \cos A} \\ &= \frac{(\cos A - \sin A)(\cos A + \sin A)}{(\cos A + \sin A)^{2}} && [\text{num: difference of two squares; denom: } (\sin A + \cos A)^{2} = 1 + 2\sin A\cos A] \\ &= \frac{\cos A - \sin A}{\cos A + \sin A} \\ &= \text{RHS} \qquad\blacksquare \end{aligned}$$LHS​=1+2sinAcosAcos2A−sin2A​=(cosA+sinA)2(cosA−sinA)(cosA+sinA)​=cosA+sinAcosA−sinA​=RHS■​​[num: difference of two squares; denom: (sinA+cosA)2=1+2sinAcosA]​

Note: the denominator 1 + 2sin A cos A = sin²A + cos²A + 2sin A cos A = (sin A + cos A)² = (cos A + sin A)².

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Common Mistakes

1. Working on both sides simultaneously — You must work from one side only.
2. Assuming the result to prove it — Do not write "LHS = RHS, therefore..." at the start.
3. Forgetting domains — Identities hold for all values in the domain. Be aware that expressions like tan θ or sec θ are undefined at certain values.
4. Incomplete simplification — Always simplify fully until you reach the exact form of the other side.

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Synoptic Links

This topic connects to:

• Reciprocal trig functions (aqa-alevel-maths-trigonometry-depth / reciprocal-trigonometric-functions) — sec²θ = 1 + tan²θ and cosec²θ = 1 + cot²θ are the workhorse identities.
• Proof by deduction (aqa-alevel-maths-pure-1 / proof) — same logical discipline applies: work one side only, justify every step.
• Integration by inspection (aqa-alevel-maths-calculus-applications / integration-trigonometric) — rewriting integrands via identity is the first move before integrating.

Summary

• An identity is true for all values of the variable in its domain.
• Work from one side only — usually the more complex side.
• Convert to sin and cos when stuck.
• Look for Pythagorean identity substitutions.
• Factorise, find common denominators, or multiply by conjugates as needed.
• State clearly which identity or technique you are using at each step.

Exam Tip: In the exam, always show every step of your working. Write "using sin²θ + cos²θ ≡ 1" or "using double angle formula" to make your reasoning transparent to the examiner. If you are stuck, try converting everything to sin and cos — this almost always gives progress. If the expression involves fractions, combining them over a common denominator is usually the right move.

A-Level Deep Dive: Proving Trigonometric Identities

Spec mapping

AQA 7357 specification, Paper 2 — Pure Mathematics, Section E (Trigonometry) covers the definitions of sine, cosine and tangent for all arguments; the sine and cosine rules; the area of a triangle in the form $\tfrac{1}{2}ab\sin C$21​absinC. Understand and use $\sin^2\theta + \cos^2\theta = 1$sin2θ+cos2θ=1, $\sec^2\theta = 1 + \tan^2\theta$sec2θ=1+tan2θ, $\csc^2\theta = 1 + \cot^2\theta$csc2θ=1+cot2θ and $\tan\theta = \sin\theta/\cos\theta$tanθ=sinθ/cosθ (refer to the official specification document for exact wording). Identity proof is examined explicitly through "Show that …" and "Prove that …" prompts on Paper 2, and is a prerequisite skill for Section E sub-strands on compound and double angles, harmonic form $R\sin(\theta + \alpha)$Rsin(θ+α), and the integration of $\sin^2 x$sin2x and $\cos^2 x$cos2x in Section H. The AQA formula booklet lists the compound-angle and double-angle formulae but does not list the Pythagorean identities — these must be memorised.

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Worked example with full mark scheme

Question (8 marks):

(a) Prove that $(1 - \cos^2\theta)(1 + \cot^2\theta) \equiv 1$(1−cos2θ)(1+cot2θ)≡1, stating any restrictions on $\theta$θ. (5)

(b) Hence solve $(1 - \cos^2\theta)(1 + \cot^2\theta) = \sin\theta + \tfrac{1}{2}$(1−cos2θ)(1+cot2θ)=sinθ+21​ for $0 \leq \theta < 2\pi$0≤θ<2π. (3)

Solution with mark scheme:

(a) Step 1 — choose a side and commit. Begin with the LHS and work toward the RHS. Never manipulate both sides simultaneously — this conflates "if" with "if and only if" and is technically invalid as a proof.

Step 2 — apply the Pythagorean identity to the first bracket.

$1 - \cos^2\theta = \sin^2\theta$1−cos2θ=sin2θ

M1 — recognising and applying $\sin^2\theta + \cos^2\theta = 1$sin2θ+cos2θ=1 rearranged.

Step 3 — apply the derived identity to the second bracket.

$1 + \cot^2\theta = \csc^2\theta = \dfrac{1}{\sin^2\theta}$1+cot2θ=csc2θ=sin2θ1​

M1 — applying $1 + \cot^2\theta = \csc^2\theta$1+cot2θ=csc2θ, then converting to $1/\sin^2\theta$1/sin2θ via the reciprocal definition.

Step 4 — combine.

$\text{LHS} = \sin^2\theta \cdot \dfrac{1}{\sin^2\theta} = 1 = \text{RHS}$LHS=sin2θ⋅sin2θ1​=1=RHS

A1 — correct cancellation reaching the RHS.

Step 5 — state the restriction.

The proof requires $\sin\theta \neq 0$sinθ=0 (so $\cot\theta$cotθ and $\csc\theta$cscθ are defined). Hence $\theta \neq k\pi$θ=kπ for integer $k$k.

B1 — explicit domain restriction stated. Many candidates omit this and lose the final mark.

A1 — proof presented as a one-sided derivation with $\equiv$≡ used correctly throughout (the question asks for an identity, not an equation).

(b) Step 1 — substitute the identity. From part (a), the LHS equals $1$1 (where defined), so:

$1 = \sin\theta + \dfrac{1}{2} \implies \sin\theta = \dfrac{1}{2}$1=sinθ+21​⟹sinθ=21​

M1 — using the result of (a) rather than re-deriving.

Step 2 — solve in the given range.

$\sin\theta = \tfrac{1}{2}$sinθ=21​ gives $\theta = \pi/6$θ=π/6 (principal value) and $\theta = \pi - \pi/6 = 5\pi/6$θ=π−π/6=5π/6 (supplementary).

A1 — both values found.

Step 3 — check the domain restriction. Both $\pi/6$π/6 and $5\pi/6$5π/6 have $\sin\theta = \tfrac{1}{2} \neq 0$sinθ=21​=0, so both are valid (the identity from (a) holds at these points).

A1 — restriction acknowledged and solutions verified.

Total: 8 marks (M3 A3 B1 plus presentation A1).

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Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks, AO2-heavy): Show that

$\dfrac{\sec\theta - \cos\theta}{\tan\theta} \equiv \sin\theta$tanθsecθ−cosθ​≡sinθ

stating any values of $\theta$θ in $[0, 2\pi)$[0,2π) for which the identity is undefined.

Mark scheme decomposition by AO:

• M1 (AO2.1) — replacing $\sec\theta$secθ with $1/\cos\theta$1/cosθ to obtain a common denominator on the numerator: $\dfrac{1}{\cos\theta} - \cos\theta = \dfrac{1 - \cos^2\theta}{\cos\theta}$cosθ1​−cosθ=cosθ1−cos2θ​.
• M1 (AO2.1) — applying the Pythagorean identity to give $\dfrac{\sin^2\theta}{\cos\theta}$cosθsin2θ​.
• M1 (AO1.1b) — replacing $\tan\theta$tanθ in the denominator by $\sin\theta/\cos\theta$sinθ/cosθ, so the full expression becomes $\dfrac{\sin^2\theta/\cos\theta}{\sin\theta/\cos\theta}$sinθ/cosθsin2θ/cosθ​.
• A1 (AO1.1b) — simplifying via division of fractions to $\sin\theta$sinθ.
• B1 (AO2.4) — restrictions: $\cos\theta \neq 0$cosθ=0 (from $\sec\theta$secθ and division by $\cos\theta$cosθ) and $\tan\theta \neq 0$tanθ=0 (denominator), giving $\theta \neq 0, \pi/2, \pi, 3\pi/2$θ=0,π/2,π,3π/2 in the given range.
• A1 (AO2.5) — proof presented as a clean one-sided derivation, ending at $\sin\theta$sinθ, with $\equiv$≡ used (not $=$=).

Total: 6 marks split AO1 = 2, AO2 = 4. This is an AO2-dominated question — AQA uses identity-proof questions specifically to test mathematical reasoning, with AO1 marks reserved for the routine algebraic manipulation. The "stating any values … undefined" rider has become a near-standard feature of AQA Paper 2 identity questions.

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Synoptic links

Connects to:

1. Compound and double-angle formulae (Section E): identities such as $\sin 2\theta = 2\sin\theta\cos\theta$sin2θ=2sinθcosθ and $\cos 2\theta = 1 - 2\sin^2\theta$cos2θ=1−2sin2θ are derived from the Pythagorean identity combined with the compound-angle expansions. Proving $\cos 2\theta = 2\cos^2\theta - 1$cos2θ=2cos2θ−1 from $\cos 2\theta = \cos^2\theta - \sin^2\theta$cos2θ=cos2θ−sin2θ uses $\sin^2\theta = 1 - \cos^2\theta$sin2θ=1−cos2θ — the same Pythagorean substitution, in a synoptic dress.

2. Harmonic form $R\sin(\theta + \alpha)$Rsin(θ+α) (Section E): expressing $a\sin\theta + b\cos\theta$asinθ+bcosθ as $R\sin(\theta + \alpha)$Rsin(θ+α) uses the compound-angle formula and Pythagoras ($R = \sqrt{a^2 + b^2}$R=a2+b2​). The harmonic form is an identity proof in disguise — you are showing that two trigonometric expressions are equivalent for all $\theta$θ.

3. Integration of $\sin^2 x$sin2x and $\cos^2 x$cos2x (Section H): the integrals $\int \sin^2 x \, dx$∫sin2xdx and $\int \cos^2 x \, dx$∫cos2xdx are inaccessible without the identity $\sin^2 x = \tfrac{1}{2}(1 - \cos 2x)$sin2x=21​(1−cos2x), which is itself a rearrangement of the double-angle identity $\cos 2x = 1 - 2\sin^2 x$cos2x=1−2sin2x. Identity manipulation is the gateway to a whole class of integrals.

4. Trigonometric equations (Section E): equations like $2\sin^2\theta + 3\cos\theta = 3$2sin2θ+3cosθ=3 are solved by replacing $\sin^2\theta$sin2θ with $1 - \cos^2\theta$1−cos2θ to convert to a quadratic in $\cos\theta$cosθ. The substitution itself is an identity proof; the rest is algebra.

5. Complex numbers and Euler's formula: $e^{i\theta} = \cos\theta + i\sin\theta$eiθ=cosθ+isinθ implies $|e^{i\theta}|^2 = \cos^2\theta + \sin^2\theta = 1$∣eiθ∣2=cos2θ+sin2θ=1 — the Pythagorean identity emerges as a single-line consequence of $|e^{i\theta}| = 1$∣eiθ∣=1. This is the deep "why" behind the identity and is often raised at Oxbridge interview.

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Mark-scheme literacy

Identity-proof questions on AQA 7357 Paper 2 split AO marks heavily toward AO2: