Cambridge A Level Mathematics 9709 · Pure 1

Trigonometric identities & equations

Cambridge 9709 Pure 1 expects exactly two trigonometric identities:

tan⁡θ=sin⁡θcos⁡θsin⁡2θ+cos⁡2θ=1\tan\theta = \frac{\sin\theta}{\cos\theta} \qquad\qquad \sin^2\theta + \cos^2\theta = 1

Everything the paper asks — proving given identities, solving equations across a range — is these two facts plus algebra. The marks are lost to two specific habits: dividing an equation by a trig function (which silently deletes solutions) and stopping at one solution when the range holds two or three. Both appear by name in Cambridge examiner reports, session after session.

Updated 20 August 2026

Using the two identities

Each identity has one standard job:

  • tan⁡θ=sin⁡θ/cos⁡θ\tan\theta = \sin\theta/\cos\theta converts a mixed sin⁡\sin/cos⁡\cos equation into one function. Seeing 3sin⁡θ=2cos⁡θ3\sin\theta = 2\cos\theta should trigger “divide by cos⁡θ\cos\theta” → tan⁡θ=23\tan\theta = \tfrac{2}{3}. (Dividing by cos⁡θ\cos\theta is safe here because cos⁡θ=0\cos\theta = 0 doesn’t satisfy the original equation — see the mistakes section for when it isn’t.)
  • sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 swaps a squared function for the other one, turning equations like 2cos⁡2θ+sin⁡θ=12\cos^2\theta + \sin\theta = 1 into a quadratic in sin⁡θ\sin\theta.

Solving equations across a range

A calculator gives one angle — sin⁡−1,cos⁡−1,tan⁡−1\sin^{-1}, \cos^{-1}, \tan^{-1} return the principal value. The range (usually 0∘≤θ≤360∘0^\circ \le \theta \le 360^\circ) holds more. Where each function repeats:

  • sin: second solution at 180∘−θ180^\circ - \theta
  • cos: second solution at 360∘−θ360^\circ - \theta
  • tan: repeats every 180∘180^\circ

A sketch of the curve — or the CAST quadrant diagram, whichever you think in — takes ten seconds and shows exactly how many answers the range holds. Count them before solving; then you know when you’re done.

Worked example

Worked example

Solve 2cos²θ + sinθ = 1 for 0° ≤ θ ≤ 360°.

Step 1 — one function. Replace cos⁡2θ\cos^2\theta using the identity:

2(1−sin⁡2θ)+sin⁡θ=1  ⇒  2sin⁡2θ−sin⁡θ−1=02(1-\sin^2\theta) + \sin\theta = 1 \;\Rightarrow\; 2\sin^2\theta - \sin\theta - 1 = 0

Step 2 — factorise the quadratic (in s=sin⁡θs = \sin\theta): (2s+1)(s−1)=0(2s+1)(s-1) = 0, so sin⁡θ=−12\sin\theta = -\tfrac12 or sin⁡θ=1\sin\theta = 1.

Step 3 — all solutions in range. sin⁡θ=1\sin\theta = 1 gives θ=90∘\theta = 90^\circ. sin⁡θ=−12\sin\theta = -\tfrac12: principal value −30∘-30^\circ is outside the range; sine is negative in the third and fourth quadrants, giving 180∘+30∘=210∘180^\circ + 30^\circ = 210^\circ and 360∘−30∘=330∘360^\circ - 30^\circ = 330^\circ.

Answer: θ=90∘,210∘,330∘\theta = 90^\circ, 210^\circ, 330^\circ — three solutions. Two of the three marks here are for the second and third; the candidate who stops at 90° scored one.

Proving identities

“Show that” questions want a chain of equalities from one side to the other. The working rules:

  1. Start from one side (usually the messier one) and transform it — don’t manipulate both sides at once as if solving an equation.
  2. Convert tan⁡\tan to sin⁡/cos⁡\sin/\cos early; combine fractions over a common denominator; deploy sin⁡2+cos⁡2=1\sin^2 + \cos^2 = 1 whenever a 11, sin⁡2\sin^2 or cos⁡2\cos^2 needs to become the others.
  3. End by writing the target expression exactly — the final line is the mark.

The mistakes that lose the marks

  1. 1.Dividing the equation by sinθ or cosθ

    sin⁡θcos⁡θ=sin⁡θ\sin\theta\cos\theta = \sin\theta divided by sin⁡θ\sin\theta loses every solution of sin⁡θ=0\sin\theta = 0. Move everything to one side and factorise instead: sin⁡θ(cos⁡θ−1)=0\sin\theta(\cos\theta - 1) = 0 keeps both families.

  2. 2.Stopping at the calculator's answer

    The inverse functions return one principal value; the range usually holds two or three solutions. Sketch or CAST first, count the crossings, and only stop when you have that many.

  3. 3.Working in degrees when the question is in radians

    If the range reads 0≤θ≤2π0 \le \theta \le 2\pi, answer in radians — exact multiples of π\pi where possible. Check the calculator mode matches before the first keystroke.

  4. 4.Discarding the 'impossible' root without saying why

    When a quadratic gives sin⁡θ=1.4\sin\theta = 1.4, reject it in writing(“no solutions since −1≤sin⁡θ≤1-1 \le \sin\theta \le 1”) — the rejection is part of the argument, and silent omission looks like an error.

  5. 5.Treating an identity proof like an equation

    You can’t “do the same to both sides” of something you’re trying to prove. Transform one side until it becomes the other; each line must follow from the last.

Common questions

What trig identities do you need for Pure 1?

Two: tan⁡θ=sin⁡θ/cos⁡θ\tan\theta = \sin\theta/\cos\theta and sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 — plus the exact values for 30°, 45°, 60° and the shapes of the three graphs. The addition formulae, double angles and Rsin⁡(θ+α)R\sin(\theta+\alpha) are Pure 3.

How many solutions should a trig equation have?

Depends on the function and the range: over 0°–360°, sin⁡θ=k\sin\theta = k or cos⁡θ=k\cos\theta = k (with ∣k∣<1|k|<1) each give two; tan⁡θ=k\tan\theta = k gives two; equations that factorise into two conditions can give up to four or five. The sketch tells you the count before you solve.

What is the CAST diagram?

A quadrant memory aid: going anticlockwise from the fourth quadrant, Cos, All, Sin, Tan are positive in those quadrants. It answers “where is the second solution?” without memorising the 180°−θ / 360°−θ rules — use whichever picture you trust under pressure.

Practise trigonometric identities & equations against real mark schemes

Quanta has real Cambridge A Level Maths 9709 past-paper questions, broken into skill checkpoints, marked criterion by criterion the way examiners mark — and it tracks which skills you’re missing. Free for individual students.

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