# How do you show that cos(2x) = 2cos^2x- 1?

Nov 10, 2016

$\cos \left(2 x\right) = \cos \left(x + x\right)$

By the sum and difference identities, we have that $\cos \left(A + B\right) = \cos A \cos B - \sin A \sin B$.

$\cos \left(x + x\right) = \cos x \cos x - \sin x \sin x$

$\cos \left(2 x\right) = {\cos}^{2} x - {\sin}^{2} x$

Use the identity ${\sin}^{2} \theta + {\cos}^{2} \theta = 1 \to {\sin}^{2} \theta = 1 - {\cos}^{2} \theta$.

$\cos \left(2 x\right) = {\cos}^{2} x - \left(1 - {\cos}^{2} x\right)$

$\cos \left(2 x\right) = {\cos}^{2} x - 1 + {\cos}^{2} x$

$\cos \left(2 x\right) = 2 {\cos}^{2} x - 1$

$L H S = R H S$

This identity has been proved!

Hopefully this helps!