3.310. $\cos 2 \alpha-\sin 4 \alpha-\cos 6 \alpha$.
## Solution.

$\cos 2 \alpha-\sin 4 \alpha-\cos 6 \alpha=(\cos 2 \alpha-\cos 6 \alpha)-\sin 4 \alpha=$ $=\left[\cos x-\cos y=-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}\right]=-2 \sin 4 \alpha \sin (-2 \alpha)-\sin 4 \alpha=$ $=2 \sin 4 \alpha \sin 2 \alpha-\sin 4 \alpha=2 \sin 4 \alpha\left(\sin 2 \alpha-\frac{1}{2}\right)=$

$=2 \sin 4 \alpha\left(\sin 2 \alpha-\sin 30^{\circ}\right)=$

$=\left[\sin x-\sin y=2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}\right]=$

$=2 \sin 4 \alpha \cdot 2 \cos \left(\alpha+15^{\circ}\right) \sin \left(\alpha-15^{\circ}\right)=$

$=4 \sin 4 \alpha \sin \left(\alpha-15^{\circ}\right) \cos \left(\alpha+15^{\circ}\right)$.

Answer: $4 \sin 4 \alpha \sin \left(\alpha-15^{\circ}\right) \cos \left(\alpha+15^{\circ}\right)$.