Putting \(B=A\) in the compound-angle formulas produces the double-angle formulas — no new memorising is needed if you can expand \(\sin(A+A)\), \(\cos(A+A)\) and \(\tan(A+A)\).
Double-angle formulas
| Formula | From \(B=A\) |
|---|---|
| \(\sin 2A=2\sin A\cos A\) | \(\sin A\cos A+\cos A\sin A\) |
| \(\cos 2A=\cos^2 A-\sin^2 A\) \(=2\cos^2 A-1\) \(=1-2\sin^2 A\) | \(\cos A\cos A-\sin A\sin A\), then replace \(\sin^2 A\) or \(\cos^2 A\) using \(\sin^2 A+\cos^2 A=1\). |
| \(\tan 2A=\dfrac{2\tan A}{1-\tan^2 A}\) | \(\dfrac{\tan A+\tan A}{1-\tan A\tan A}\) |
Choosing a form of \(\cos 2A\)
| You want | Use |
|---|---|
| Sines only | \(\cos 2A=1-2\sin^2 A\) |
| Cosines only | \(\cos 2A=2\cos^2 A-1\) |
| Rearranged forms for proofs | \(1-\cos 2A=2\sin^2 A\) \(1+\cos 2A=2\cos^2 A\) |
Double of any angle
In every formula, \(2A\) means "double of the angle inside" — the formulas apply to any pair of angles in the ratio \(2:1\).
| Statement | Read as |
|---|---|
| \(\sin 6\theta=2\sin 3\theta\cos 3\theta\) | \(6\theta\) is double \(3\theta\). |
| \(\sin\theta=2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}\) | \(\theta\) is double \(\dfrac{\theta}{2}\). |
| \(\cos 8\theta=1-2\sin^2 4\theta\) | \(8\theta\) is double \(4\theta\). |
Doubling the angle does not double the ratio: \(\cos 2A\ne 2\cos A\) and \(\sin 2A\ne 2\sin A\). Always expand with the formulas.
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