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
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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 $$\cos^2 \theta$$.

Proof of Corollary
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Alternative Proof
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