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TMUA Trigonometric Equations and Identities

Trigonometric equations ask for all angles in a stated interval. You need exact values, the two main identities and a careful count of solutions.

Paper 1 and Paper 2 topic. Specification codes: MM4.3, MM4.4, MM4.5, MM4.6. Last checked 12 November 2026.

Identities and exact values

  • tan⁡θ=sin⁡θcos⁡θ\tan\theta=\dfrac{\sin\theta}{\cos\theta} and sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1.
  • Know sin⁡\sin, cos⁡\cos and tan⁡\tan of 0∘0^\circ, 30∘30^\circ, 45∘45^\circ, 60∘60^\circ and 90∘90^\circ exactly.

Finding every solution

Find the principal value, then use the symmetry of the graph for the others: sin⁡θ=sin⁡(180∘−θ)\sin\theta=\sin(180^\circ-\theta), cos⁡θ=cos⁡(360∘−θ)\cos\theta=\cos(360^\circ-\theta), and add 180∘180^\circ for tangent.

Multiple angles

For sin⁡2x\sin 2x, first solve for 2x2x over a doubled interval, then halve. Count the solutions before you halve.

Graph shapes and symmetries

Sine and cosine repeat every 360∘360^\circ (2π2\pi) and tangent every 180∘180^\circ (π\pi). Also sin⁡(−θ)=−sin⁡θ\sin(-\theta)=-\sin\theta and cos⁡(−θ)=cos⁡θ\cos(-\theta)=\cos\theta. A quick sketch tells you how many solutions to expect.

If an equation is a quadratic in sin⁡x\sin x or cos⁡x\cos x, factorise first. A value outside −1-1 to 11 gives no solutions. Never cancel a factor of sin⁡x\sin x or cos⁡x\cos x without checking it could be zero.

Always read the required interval carefully. Solutions in 0≤x≤2π0\le x\le2\pi are not the same as solutions in −π≤x≤π-\pi\le x\le\pi, and the end values are included only when the inequality says so.

Worked examples

Example 1

Solve 2sin⁡2x=1+cos⁡x2\sin^2x=1+\cos x for 0∘≤x≤360∘0^\circ\le x\le360^\circ.

Show the worked solution

Use sin⁡2x=1−cos⁡2x\sin^2x=1-\cos^2x: 2−2cos⁡2x=1+cos⁡x2-2\cos^2x=1+\cos x, so 2cos⁡2x+cos⁡x−1=02\cos^2x+\cos x-1=0, i.e. (2cos⁡x−1)(cos⁡x+1)=0(2\cos x-1)(\cos x+1)=0. Then cos⁡x=12\cos x=\tfrac12 gives 60∘,300∘60^\circ,300^\circ, and cos⁡x=−1\cos x=-1 gives 180∘180^\circ.

Answer: x=60∘, 180∘, 300∘x=60^\circ,\,180^\circ,\,300^\circ

Example 2

How many solutions does sin⁡2x=12\sin2x=\tfrac12 have for 0∘≤x<360∘0^\circ\le x<360^\circ?

Show the worked solution

Now 0∘≤2x<720∘0^\circ\le2x<720^\circ. The values of 2x2x are 30∘,150∘,390∘,510∘30^\circ,150^\circ,390^\circ,510^\circ, so x=15∘,75∘,195∘,255∘x=15^\circ,75^\circ,195^\circ,255^\circ.

Answer: 4

Example 3

Given tan⁡θ=2\tan\theta=2 with θ\theta acute, find sin⁡θcos⁡θ\sin\theta\cos\theta.

Show the worked solution

The triangle has sides 1,2,51,2,\sqrt5. So sin⁡θ=25\sin\theta=\tfrac2{\sqrt5}, cos⁡θ=15\cos\theta=\tfrac1{\sqrt5} and the product is 25\tfrac25.

Answer: 25\tfrac25

Common mistakes

  • Giving only the principal value and missing the other solutions in the interval.
  • Halving the interval for sin⁡2x\sin2x instead of doubling it.
  • Dividing both sides by cos⁡x\cos x and losing the solutions where cos⁡x=0\cos x=0.
  • Mixing degrees and radians.

Test yourself on this topic with our free practice questions.

Quick answers

How do you solve a trigonometric equation for all solutions?

Find the principal value, use the symmetry of the graph for the other values, then check each lies in the given interval.

What is sin squared plus cos squared?

sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 for every angle θ\theta.

Do I need to memorise exact trig values?

Yes for 0∘0^\circ, 30∘30^\circ, 45∘45^\circ, 60∘60^\circ and 90∘90^\circ. There is no calculator to give them to you.

Practise it under timed conditions

A free timed sample paper with a worked solution for every question. No sign-up. For two full papers and a weak-spot drill, see Full Access.

Free sample paper

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