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

$$
\begin{aligned}
& \cos 2 \alpha+\sin 4 \alpha-\cos 6 \alpha=(\cos 2 \alpha-\cos 6 \alpha)+\sin 4 \alpha= \\
& =\left[\cos x-\cos y=-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}\right]=-2 \sin 4 \alpha \sin (-2 \alpha)+\sin 4 \alpha= \\
& =2 \sin 4 \alpha \sin 2 \alpha+\sin 4 \alpha=2 \sin 4 \alpha\left(\sin 2 \alpha+\frac{1}{2}\right)= \\
& =2 \sin 4 \alpha\left(\sin 2 \alpha+\sin 30^{\circ}\right)=\left[\sin x+\sin y=2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}\right]= \\
& =2 \sin 4 \alpha \cdot 2 \sin \left(\alpha+15^{\circ}\right) \cos \left(\alpha-15^{\circ}\right)=4 \sin 4 \alpha \sin \left(\alpha+15^{\circ}\right) \cos \left(\alpha-15^{\circ}\right)
\end{aligned}
$$

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