## Precalculus (6th Edition)

$\sin^4\theta-\cos^4\theta=2\sin^2\theta-1$
Start with the left side: $\sin^4\theta-\cos^4\theta$ Write it as a difference of perfect squares and factor: $=(\sin^2\theta)^2-(\cos^2\theta)^2$ $=(\sin^2\theta+\cos^2\theta)(\sin^2\theta-\cos^2\theta)$ Simplify: $=1*(\sin^2\theta-\cos^2\theta)$ $=\sin^2\theta-\cos^2\theta$ $=\sin^2\theta-(1-\sin^2\theta)$ $=\sin^2\theta-1+\sin^2\theta$ $=2\sin^2\theta-1$ Since this equals the right side, the identity has been proven.