Trigonometric Identities

Trigonometric identities are equations that are true for all values of the variable (where both sides are defined). At A-Level (9MA0), you must know the fundamental identities, use them to simplify expressions, prove further identities, and solve equations.

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The Two Fundamental Identities

Identity 1: tan(theta) = sin(theta) / cos(theta)

This follows directly from the definitions. It is valid for all theta where cos(theta) ≠ 0.

Identity 2: sin²(theta) + cos²(theta) = 1

This is the Pythagorean identity. It holds for all values of theta.

Rearrangements:

• sin²(theta) = 1 - cos²(theta)
• cos²(theta) = 1 - sin²(theta)

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Proving Identities

When proving a trigonometric identity, work with one side (usually the more complicated side) and show it simplifies to the other.

Golden rules:

1. Start with one side only — do NOT work both sides simultaneously.
2. Replace tan with sin/cos when useful.
3. Use sin² + cos² = 1 to eliminate one trig function.
4. Factorise where possible.

Worked Example 1

Prove that (1 - cos²x) / sin(x) = sin(x).

Start with the left side:

LHS = (1 - cos²x) / sin(x)

Since 1 - cos²x = sin²x:

= sin²x / sin(x)

= sin(x)

= RHS

Worked Example 2

Prove that tan²(theta) + 1 = 1/cos²(theta).

LHS = sin²(theta)/cos²(theta) + 1

= sin²(theta)/cos²(theta) + cos²(theta)/cos²(theta)

= (sin²(theta) + cos²(theta)) / cos²(theta)

= 1/cos²(theta)

= RHS

This result is sometimes written as sec²(theta) = 1 + tan²(theta) (covered later in the course).

Worked Example 3

Prove that (sin(x) + cos(x))² = 1 + 2sin(x)cos(x).

LHS = sin²(x) + 2sin(x)cos(x) + cos²(x)

= (sin²(x) + cos²(x)) + 2sin(x)cos(x)

= 1 + 2sin(x)cos(x)

= RHS

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Solving Equations Using Identities

Identities allow you to write equations in terms of a single trig function, which can then be solved.

Worked Example 4

Solve 2sin²(x) + cos(x) - 1 = 0 for 0 <= x < 2pi.

Replace sin²(x) with 1 - cos²(x):

2(1 - cos²(x)) + cos(x) - 1 = 0

2 - 2cos²(x) + cos(x) - 1 = 0

-2cos²(x) + cos(x) + 1 = 0

2cos²(x) - cos(x) - 1 = 0

Let c = cos(x):

2c² - c - 1 = 0

(2c + 1)(c - 1) = 0

c = -1/2 or c = 1

cos(x) = -1/2: x = 2pi/3, 4pi/3

cos(x) = 1: x = 0

Solutions: x = 0, 2pi/3, 4pi/3

Worked Example 5

Solve 3sin(x) = 2tan(x) for 0 <= x < 2pi.

Replace tan(x) = sin(x)/cos(x):

3sin(x) = 2sin(x)/cos(x)

Multiply both sides by cos(x) (noting cos(x) ≠ 0):

3sin(x)cos(x) = 2sin(x)

3sin(x)cos(x) - 2sin(x) = 0

sin(x)(3cos(x) - 2) = 0

sin(x) = 0: x = 0, pi

cos(x) = 2/3: x = cos^(-1)(2/3) ≈ 0.8411, and x = 2pi - 0.8411 ≈ 5.4421

Solutions: x = 0, 0.841, pi, 5.442 (to 3 d.p.)

Worked Example 6

Solve 4cos²(theta) - 3 = 0 for 0 <= theta <= 2pi.

cos²(theta) = 3/4

cos(theta) = ± sqrt(3)/2

cos(theta) = sqrt(3)/2: theta = pi/6, 11pi/6

cos(theta) = -sqrt(3)/2: theta = 5pi/6, 7pi/6

Solutions: theta = pi/6, 5pi/6, 7pi/6, 11pi/6

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Common Mistakes

Mistake	Correction
sin²(x) + cos²(x) = 1 written as sin(x²) + cos(x²) = 1	sin² means (sin x)², not sin(x²)
Dividing by sin(x) and losing solutions	Factor instead: sin(x)(something) = 0 gives sin(x) = 0 OR something = 0
Working both sides of an identity at once	Start with one side and transform to the other
Forgetting to check for extraneous solutions	If you multiply by cos(x), check cos(x) ≠ 0

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Exam Tips

Tip	Detail
Prove means show	You must show every step — do not skip algebra
Factor, do not divide	When sin(x) is a common factor, factorise to keep all solutions
Substitute early	Replace tan with sin/cos or use sin² + cos² = 1 early to simplify
Check solutions	Substitute back into the original equation to verify
Look for quadratics	If the equation has mixed trig functions, aim for a quadratic in one function

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Practice Problems

1. Prove that (1 + tan²x) x cos²x = 1.
2. Solve 2cos²(x) - cos(x) = 0 for 0 <= x < 2pi. [Answer: pi/3, pi/2, 3pi/2, 5pi/3]
3. Prove that sin(x)/(1 + cos(x)) + (1 + cos(x))/sin(x) = 2/sin(x).
4. Solve 5sin²(x) + 3cos(x) - 3 = 0 for 0 <= x <= 2pi. [Answer: x = pi/3, 5pi/3, pi]
5. Prove that (cos(x) - sin(x))² + (cos(x) + sin(x))² = 2.

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Summary

• Two fundamental identities: tan(theta) = sin(theta)/cos(theta) and sin²(theta) + cos²(theta) = 1.
• To prove an identity, start with one side and manipulate to reach the other.
• To solve equations, use identities to reduce to a single trig function, then solve.
• Always factor rather than divide when a trig function could be zero.

A-Level Deep Dive: Trigonometric Identities

Spec mapping

Edexcel 9MA0 specification section 5 — Trigonometry: "Understand and use the definitions of $\sin$sin, $\cos$cos and $\tan$tan for all arguments; the sine and cosine rules; the area of a triangle in the form $\tfrac{1}{2}ab\sin C$21​absinC; sine, cosine and tangent functions, their graphs, symmetries and periodicity; the Pythagorean identity $\sin^2\theta + \cos^2\theta \equiv 1$sin2θ+cos2θ≡1; the quotient identity $\tan\theta \equiv \sin\theta/\cos\theta$tanθ≡sinθ/cosθ; the secant, cosecant and cotangent functions and their relationships." The two derived Pythagorean identities $1 + \tan^2\theta \equiv \sec^2\theta$1+tan2θ≡sec2θ and $1 + \cot^2\theta \equiv \csc^2\theta$1+cot2θ≡csc2θ are named in the specification and listed in the Edexcel formula booklet under the trigonometry heading. Synoptically these basic identities underpin almost every other trig topic on Paper 2: compound-angle formulae $\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$sin(A±B)=sinAcosB±cosAsinB and the double-angle formulae that follow from them; the harmonic form $a\sin\theta + b\cos\theta = R\sin(\theta + \alpha)$asinθ+bcosθ=Rsin(θ+α); proving identities as a stand-alone question type in section 5.4; and the integration techniques of section 8 that depend on $\sin^2 x = \tfrac{1}{2}(1 - \cos 2x)$sin2x=21​(1−cos2x) and $\cos^2 x = \tfrac{1}{2}(1 + \cos 2x)$cos2x=21​(1+cos2x). The inverse trigonometric functions of section 5.6 also rely on Pythagorean identities to convert between forms (e.g. $\cos(\arcsin x) = \sqrt{1 - x^2}$cos(arcsinx)=1−x2​).

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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 for all $\theta$θ such that $\sin\theta \neq 0$sinθ=0. (4)

(b) Hence, or otherwise, solve the equation $(1 - \cos^2\theta)(1 + \cot^2\theta) = 2\sin\theta$(1−cos2θ)(1+cot2θ)=2sinθ for $0 \leq \theta < 2\pi$0≤θ<2π, giving exact answers in terms of $\pi$π. (4)

Solution with mark scheme:

(a) Step 1 — start from the LHS, replace the first bracket using the Pythagorean identity.

From $\sin^2\theta + \cos^2\theta \equiv 1$sin2θ+cos2θ≡1 we have $1 - \cos^2\theta \equiv \sin^2\theta$1−cos2θ≡sin2θ. So:

$\text{LHS} = (1 - \cos^2\theta)(1 + \cot^2\theta) = \sin^2\theta(1 + \cot^2\theta)$LHS=(1−cos2θ)(1+cot2θ)=sin2θ(1+cot2θ)

M1 — correct first substitution from a named Pythagorean identity. Examiners expect the candidate to begin from one side and transform it; substituting $1 - \cos^2\theta$1−cos2θ first is the standard route. Common error: candidates expand the brackets directly into $1 + \cot^2\theta - \cos^2\theta - \cos^2\theta\cot^2\theta$1+cot2θ−cos2θ−cos2θcot2θ and get tangled in four-term simplification — technically possible, but loses the elegance and very often loses the M-mark for "no clear progress toward RHS".

Step 2 — replace the second bracket using the derived Pythagorean identity.

The identity $1 + \cot^2\theta \equiv \csc^2\theta$1+cot2θ≡csc2θ gives:

$\sin^2\theta(1 + \cot^2\theta) = \sin^2\theta \cdot \csc^2\theta$sin2θ(1+cot2θ)=sin2θ⋅csc2θ

M1 — correct application of the second derived Pythagorean identity. Naming it explicitly in the working (e.g. "applying $1 + \cot^2\theta \equiv \csc^2\theta$1+cot2θ≡csc2θ") is the kind of phrasing that secures method credit even if the candidate writes the wrong line later.

Step 3 — use the reciprocal definition.

Since $\csc\theta = \dfrac{1}{\sin\theta}$cscθ=sinθ1​, we have $\csc^2\theta = \dfrac{1}{\sin^2\theta}$csc2θ=sin2θ1​. Substituting:

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

M1 — correct use of the reciprocal definition $\csc\theta = 1/\sin\theta$cscθ=1/sinθ.

A1 — clean reduction to $1$1, with explicit identification of RHS (the closing line "$= \text{RHS}$=RHS as required" is the rewarded phrasing).

(b) Step 1 — use the result of (a).

The LHS equals $1$1 for all valid $\theta$θ. So the equation reduces to:

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

M1 — using the proven identity to simplify, honouring the "Hence" command.

Step 2 — solve in the given interval.

The principal value: $\theta = \dfrac{\pi}{6}$θ=6π​. The second solution in $[0, 2\pi)$[0,2π): by the symmetry $\sin(\pi - \theta) = \sin\theta$sin(π−θ)=sinθ, we get $\theta = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$θ=π−6π​=65π​.

M1 — correct method for finding both solutions in the interval.

A1 — first solution $\theta = \pi/6$θ=π/6.

A1 — second solution $\theta = 5\pi/6$θ=5π/6.

Check the validity constraint: both solutions satisfy $\sin\theta \neq 0$sinθ=0, so neither is excluded by the domain restriction in (a).

Total: 8 marks (M4 A4, split as shown).

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Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): Prove that $\dfrac{1}{1 - \sin\theta} + \dfrac{1}{1 + \sin\theta} \equiv 2\sec^2\theta$1−sinθ1​+1+sinθ1​≡2sec2θ for all $\theta$θ with $\cos\theta \neq 0$cosθ=0.

Mark scheme decomposition by AO:

• M1 (AO1.1a) — combining the two fractions over a common denominator: $\dfrac{(1 + \sin\theta) + (1 - \sin\theta)}{(1 - \sin\theta)(1 + \sin\theta)}$(1−sinθ)(1+sinθ)(1+sinθ)+(1−sinθ)​.
• A1 (AO1.1b) — simplifying the numerator to $2$2 and the denominator to $1 - \sin^2\theta$1−sin2θ via difference of squares.
• M1 (AO1.1b) — recognising $1 - \sin^2\theta \equiv \cos^2\theta$1−sin2θ≡cos2θ from the Pythagorean identity and substituting.
• A1 (AO1.1b) — writing the result as $\dfrac{2}{\cos^2\theta}$cos2θ2​.
• M1 (AO2.1) — using the reciprocal definition $\sec\theta = 1/\cos\theta$secθ=1/cosθ to rewrite as $2\sec^2\theta$2sec2θ.
• A1 (AO2.5) — clean conclusion "$\equiv \text{RHS}$≡RHS as required" with the domain restriction $\cos\theta \neq 0$cosθ=0 noted.

Total: 6 marks split AO1 = 4, AO2 = 2. Identity-proof questions on Paper 2 are AO1-dominated (procedural fluency in identity manipulation) but reserve AO2 marks for the strategic moves: choosing the right Pythagorean form, recognising the difference-of-squares pattern, and finishing with explicit reciprocal definitions rather than mid-line rewrites.

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Synoptic links

Connects to:

1. Section 5.5 — Compound angle and double angle formulae: $\sin(A + B) = \sin A\cos B + \cos A\sin B$sin(A+B)=sinAcosB+cosAsinB and $\cos 2A = 1 - 2\sin^2 A = 2\cos^2 A - 1$cos2A=1−2sin2A=2cos2A−1 are derived from and constantly use the Pythagorean identity. The double-angle forms for $\cos 2A$cos2A are exactly $\cos^2 A - \sin^2 A$cos2A−sin2A rewritten via $\sin^2 A + \cos^2 A = 1$sin2A+cos2A=1. Without confident Pythagorean manipulation, the double-angle conversions stall.

2. Section 5.7 — Harmonic form $R\sin(\theta + \alpha)$Rsin(θ+α): expressing $a\sin\theta + b\cos\theta$asinθ+bcosθ in the form $R\sin(\theta + \alpha)$Rsin(θ+α) with $R = \sqrt{a^2 + b^2}$R=a2+b2​ uses $\sin^2\alpha + \cos^2\alpha = 1$sin2α+cos2α=1 at the heart of the derivation: setting $R\cos\alpha = a$Rcosα=a and $R\sin\alpha = b$Rsinα=b, squaring and adding gives $R^2 = a^2 + b^2$R2=a2+b2 exactly because of the Pythagorean identity.

3. Section 8 — Integration of trigonometric powers: integrals like $\int \sin^2 x\,dx$∫sin2xdx and $\int \cos^4 x\,dx$∫cos4xdx are routinely tackled by replacing $\sin^2 x = \tfrac{1}{2}(1 - \cos 2x)$sin2x=21​(1−cos2x), an identity that follows from the Pythagorean identity combined with the double-angle form $\cos 2x = 1 - 2\sin^2 x$cos2x=1−2sin2x. Integration of $\sec^2\theta\tan\theta$sec2θtanθ also leans on $1 + \tan^2\theta = \sec^2\theta$1+tan2θ=sec2θ.

4. Section 5.4 — Trigonometric equations: solving $2\sin^2\theta + 3\cos\theta - 3 = 0$2sin2θ+3cosθ−3=0 requires replacing $\sin^2\theta$sin2θ with $1 - \cos^2\theta$1−cos2θ to produce a quadratic in $\cos\theta$cosθ. Almost every "mixed-function" trig equation reduces to a single function via Pythagorean substitution before factorising.

5. Year-13 Further Maths — Complex numbers and Euler's formula: $e^{i\theta} = \cos\theta + i\sin\theta$eiθ=cosθ+isinθ encodes the Pythagorean identity as $|e^{i\theta}|^2 = \cos^2\theta + \sin^2\theta = 1$∣eiθ∣2=cos2θ+sin2θ=1, the statement that the unit circle in the complex plane has unit modulus. Hyperbolic identities $\cosh^2 x - \sinh^2 x = 1$cosh2x−sinh2x=1 are the analogue for the unit hyperbola, derived in exactly the same algebraic way.

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Mark-scheme literacy

"Prove that …" identity questions on 9MA0 split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	60–70%	Applying $\sin^2 + \cos^2 \equiv 1$sin2+cos2≡1, $1 + \tan^2 \equiv \sec^2$1+tan2≡sec2, $1 + \cot^2 \equiv \csc^2$1+cot2≡csc2; using quotient and reciprocal definitions; combining algebraic fractions cleanly
AO2 (reasoning / interpretation)	25–35%	Choosing the correct side to start from; selecting the most efficient identity at each step; presenting the proof as a chain of equivalences; stating domain restrictions where reciprocals appear
AO3 (problem-solving)	0–10%	Open-ended modelling rare for this topic; appears in synoptic harmonic-form / mechanics contexts

Examiner-rewarded phrasing: "starting from the LHS, applying $\sec^2\theta \equiv 1 + \tan^2\theta$sec2θ≡1+tan2θ"; "since $\cos\theta \neq 0$cosθ=0 in the given interval, division by $\cos\theta$cosθ is valid"; "combining over the common denominator $\sin\theta\cos\theta$sinθcosθ"; "= RHS as required". Phrases that lose marks: working both sides of the identity simultaneously and "meeting in the middle" — examiners often reject this as a logical structure because it can hide circular reasoning. Writing "$= \text{LHS}$=LHS" at the start and "$= \text{RHS}$=RHS" at the end of a chain that should be $\text{LHS} = \dots = \text{RHS}$LHS=⋯=RHS is the formal expectation.

A specific 9MA0 pattern to watch: questions phrased "Prove that $f(\theta) \equiv g(\theta)$f(θ)≡g(θ)" use the identity symbol $\equiv$≡ deliberately. The proof must hold for all $\theta$θ in the stated domain, not just the values that make both sides defined accidentally. State the excluded points (e.g. "for $\theta \neq \pi/2 + k\pi$θ=π/2+kπ, where $\sec\theta$secθ is undefined") explicitly.

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Grade-band model answers

3-mark question

Question: Prove that $\sin\theta\cot\theta \equiv \cos\theta$sinθcotθ≡cosθ for all $\theta$θ with $\sin\theta \neq 0$sinθ=0.

Grade C response (~210 words):

Starting from the LHS:

$\sin\theta\cot\theta = \sin\theta \cdot \dfrac{\cos\theta}{\sin\theta}$sinθcotθ=sinθ⋅sinθcosθ​.

The $\sin\theta$sinθ cancels:

$= \cos\theta = \text{RHS}$=cosθ=RHS.

So $\sin\theta\cot\theta \equiv \cos\theta$sinθcotθ≡cosθ as required.