3.075. $(\cos \alpha-\cos 2 \beta)^{2}+(\sin \alpha+\sin 2 \beta)^{2}$.
## Solution.

$(\cos \alpha-\cos 2 \beta)^{2}+(\sin \alpha+\sin 2 \beta)^{2}=$

$=\cos ^{2} \alpha-2 \cos \alpha \cos 2 \beta+\cos ^{2} 2 \beta+\sin ^{2} \alpha+2 \sin \alpha \sin 2 \beta+\sin ^{2} 2 \beta=$

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

$=2-2(\cos \alpha \cos 2 \beta-\sin \alpha \sin 2 \beta)=2-2 \cos (\alpha+2 \beta)=$

$=2-2 \cos \left(2 \cdot \frac{\alpha+2 \beta}{2}\right)=2-2\left(1-2 \sin ^{2} \frac{\alpha+2 \beta}{2}\right)=2-2+4 \sin ^{2} \frac{\alpha+2 \beta}{2}=$ $=4 \sin ^{2} \frac{\alpha+2 \beta}{2}$.

Answer: $4 \sin ^{2} \frac{\alpha+2 \beta}{2}$.