# How do you verify the identity cos^4theta-sin^4theta=cos^2theta-sin^2theta?

Sep 18, 2016

Factor the left hand side as a difference of squares:

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

Apply the pythagorean identity ${\cos}^{2} \theta + {\sin}^{2} \theta = 1$:

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

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

$L H S = R H S$

Hopefully this helps!