Double Angle Formulas

Theorem

 * $\sin \left({2 \theta}\right) = 2 \sin \theta \cos \theta$
 * $\cos \left({2 \theta}\right) = \cos^2 \theta - \sin^2 \theta$
 * $\displaystyle \tan \left({2 \theta}\right) = \frac {2\tan \theta} {1 - \tan^2 \theta}$

where $\sin, \cos, \tan$ are sine, cosine and tangent.

Corollary

 * $\cos \left({2 \theta}\right) = 2 \cos^2 \theta - 1$
 * $\cos \left({2 \theta}\right) = 1 - 2 \sin^2 \theta$

Proof
We then equate real and imaginary parts:

Since $\displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta}$, we have:
 * $\displaystyle \tan \left({2 \theta}\right) = \frac{2 \sin \theta \cos \theta}{\cos^2 \theta - \sin^2 \theta}$

which is equal to:
 * $\displaystyle \frac {2 \tan \theta} {1 - \tan^2 \theta}$

by dividing the top and bottom by $\cos^2 \theta$.