Leave lesson

Pure · Trigonometry

Chapter 1 · 3

The idea

Double-angle formulae

The double-angle formulae for sin 2A, cos 2A (three forms) and tan 2A, where they come from, choosing the right form of cos 2A, and using them to solve equations and find exact values.

A full journey — read it, play with it, work it, then earn real exam marks. Everything stays on the timeline below.

In this lesson — start anywhere

Pure · Trigonometry

Double-angle formulae

The double-angle formulae for sin 2A, cos 2A (three forms) and tan 2A, where they come from, choosing the right form of cos 2A, and using them to solve equations and find exact values.

Why it works

One trick: set B = A

You don't need to memorise where these come from — you already own them. Take the addition formulae from the last lesson and make the two angles the same: put B=AB = A, and sin⁡(A+B)\sin(A + B) becomes sin⁡2A\sin 2A. Everything on this page falls out of that single move:

sin⁡(A+A)=sin⁡Acos⁡A+cos⁡Asin⁡A  ⟹  sin⁡2A≡2sin⁡Acos⁡A,\sin(A + A) = \sin A\cos A + \cos A\sin A \;\Longrightarrow\; \sin 2A \equiv 2\sin A\cos A,

and the same substitution into tan⁡(A+B)\tan(A+B) gives

tan⁡2A≡2tan⁡A1−tan⁡2A.\tan 2A \equiv \frac{2\tan A}{1 - \tan^2 A}.

Keep reading — free

The rest of the explanation, plus 3 worked examples you step through move by move.

Start free

Takes a minute — no card.