A-level Maths revision · Pure, year 2

Addition and double-angle formulae

The idea

sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\sin(A \pm B) = \sin A\cos B \pm \cos A\sin Bcos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\cos(A \pm B) = \cos A\cos B \mp \sin A\sin BWith A=BA = B: sin⁡2A=2sin⁡Acos⁡A\sin 2A = 2\sin A\cos A, and cos⁡2A\cos 2A can be written cos⁡2A−sin⁡2A\cos^2 A - \sin^2 A, 2cos⁡2A−12\cos^2 A - 1 or 1−2sin⁡2A1 - 2\sin^2 A.

  1. Split an angle into ones you know: 75∘=45∘+30∘75^\circ = 45^\circ + 30^\circ.
  2. Pick the form of cos⁡2A\cos 2A that leaves one trig function.
  3. Simplify and solve or prove.

Watch out sin⁡2A\sin 2A is 2sin⁡Acos⁡A2\sin A\cos A, not 2sin⁡A2\sin A.

See it worked

Worked example

Find the exact value of sin⁡75∘\sin 75^\circ.

  1. sin⁡(45∘+30∘)=sin⁡45∘cos⁡30∘+cos⁡45∘sin⁡30∘\sin(45^\circ + 30^\circ) = \sin 45^\circ\cos 30^\circ + \cos 45^\circ\sin 30^\circ.
  2. =22⋅32+22⋅12= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}.
  3. =6+24= \frac{\sqrt{6} + \sqrt{2}}{4}.

Answer 6+24\frac{\sqrt{6} + \sqrt{2}}{4}

Your turn

Question 1Non-calculator3 marks

Find the exact value of cos⁡15∘\cos 15^\circ.

Question 2Non-calculator3 marks

sin⁡x=35\sin x = \frac{3}{5} and xx is acute. Find the exact value of sin⁡2x\sin 2x.

Question 3Non-calculator3 marks

Solve cos⁡2x=cos⁡x\cos 2x = \cos x for 0∘≤x<360∘0^\circ \le x < 360^\circ.

In past papers

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