Chapter 7 - Trigonometric Identities and Equations - 7.4 Double-Angle and Half-Angle Identities - 7.4 Exercises - Page 694: 82

$(\cos 2x-\sin 2x)^2=1-\sin 4x$

Start with the left side: $(\cos 2x-\sin 2x)^2$ Expand: $=\cos^2 2x-2\cos 2x\sin 2x+\sin^2 2x$ Rearrange terms: $=\cos^2 2x+\sin^2 2x-2\cos 2x\sin 2x$ Use the identities $\cos^2 \theta+\sin^2\theta=1$ and $2\cos\theta\sin\theta=\sin 2\theta$, where $\theta=2x$: $=1-\sin (2*2x)$ $=1-\sin 4x$