Trigonify

Topics · 3 identities

Double Angle

Rewriting functions of 2θ in terms of functions of θ.

Double angle formula: sin 2θ = 2 sin θ cos θ

sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2 \sin\theta \cos\theta

sin 2θ = 2 sin θ cos θ, which drops straight out of the sine sum formula with β = α. It is the substitution that collapses a product of a sine and a cosine into a single term, and the one that makes ∫ sin θ cos θ dθ tractable.

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Double angle formula: cos 2θ in all three forms

cos⁡2θ=cos⁡2θ−sin⁡2θ=2cos⁡2θ−1=1−2sin⁡2θ\cos 2\theta = \cos^{2}\theta - \sin^{2}\theta = 2\cos^{2}\theta - 1 = 1 - 2\sin^{2}\theta

cos 2θ has three equivalent forms: cos²θ − sin²θ, 2cos²θ − 1, and 1 − 2sin²θ. The Pythagorean identity converts between them. Pick whichever leaves you with only the function you already have - that choice is usually what makes a problem fall out in one line instead of five.

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Double angle formula: tan 2θ = 2 tan θ / (1 − tan²θ)

tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta = \dfrac{2\tan\theta}{1 - \tan^{2}\theta}

tan 2θ = 2 tan θ / (1 − tan²θ), from the tangent sum formula with β = α. It is undefined when tan²θ = 1, i.e. at odd multiples of π/4, which is exactly where 2θ lands on an asymptote of tangent.

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