# How do you prove cos2θ = cos²θ - sin²θ?

May 23, 2015

It depends where you want to start.

If you know the Taylor series for ${e}^{z}$, $\sin \theta$ and $\cos \theta$, together with the basic properties of $i = \sqrt{- 1}$, then you can easily find that:

${e}^{i \theta} = \cos \theta + i \sin \theta$

Then:

$\cos 2 \theta + i \sin 2 \theta = {e}^{2 i \theta} = {\left({e}^{i \theta}\right)}^{2}$

$= {\left(\cos \theta + i \sin \theta\right)}^{2}$

$= {\cos}^{2} \theta + 2 i \cos \theta \sin \theta + {i}^{2} {\sin}^{2} \theta$

$= {\cos}^{2} \theta + 2 i \cos \theta \sin \theta - {\sin}^{2} \theta$

$= \left({\cos}^{2} \theta - {\sin}^{2} \theta\right) + i \left(2 \cos \theta \sin \theta\right)$

Comparing the real and imaginary parts we get two formulae for the price of one:

$\cos 2 \theta = {\cos}^{2} \theta - {\sin}^{2} \theta$

$\sin 2 \theta = 2 \cos \theta \sin \theta$