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Pure · Trigonometry

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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.

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

Put B=AB = A into the addition formulae and the compound angle becomes a double angle: sin⁡2A≡2sin⁡Acos⁡A,tan⁡2A≡2tan⁡A1−tan⁡2A.\sin 2A \equiv 2\sin A\cos A, \qquad \tan 2A \equiv \frac{2\tan A}{1 - \tan^2 A}. For cosine, cos⁡2A≡cos⁡2A−sin⁡2A\cos 2A \equiv \cos^2 A - \sin^2 A, and using sin⁡2A+cos⁡2A≡1\sin^2 A + \cos^2 A \equiv 1 this rearranges into three interchangeable forms: cos⁡2A≡cos⁡2A−sin⁡2A≡2cos⁡2A−1≡1−2sin⁡2A.\cos 2A \equiv \cos^2 A - \sin^2 A \equiv 2\cos^2 A - 1 \equiv 1 - 2\sin^2 A.

Choosing the right cos⁡2A\cos 2A form is the key skill. Pick the version that matches the other terms in the equation:
  • equation written in sin⁡\sin? use cos⁡2A=1−2sin⁡2A\cos 2A = 1 - 2\sin^2 A;
  • equation written in cos⁡\cos? use cos⁡2A=2cos⁡2A−1\cos 2A = 2\cos^2 A - 1.
That way everything becomes a quadratic in a single function.

These are not coefficients. cos⁡2A\cos 2A is not 2cos⁡A2\cos A, and sin⁡2A\sin 2A is not 2sin⁡A2\sin A — the 22 multiplies the angle, and the formula above is what that actually equals.

Solving — don't divide away a factor. For sin⁡2x=sin⁡x\sin 2x = \sin x, write 2sin⁡xcos⁡x=sin⁡x2\sin x\cos x = \sin x and factor as sin⁡x(2cos⁡x−1)=0\sin x(2\cos x - 1) = 0; dividing both sides by sin⁡x\sin x would throw away the sin⁡x=0\sin x = 0 solutions.