3.127. $1+\cos \left(2 \alpha+270^{\circ}\right)+\sin \left(2 \alpha+450^{\circ}\right)$.
## Solution.

$$
\begin{aligned}
& 1+\cos \left(2 \alpha+270^{\circ}\right)+\sin \left(2 \alpha+450^{\circ}\right)=1+\cos \left(270^{\circ}+2 \alpha\right)+\sin \left(450^{\circ}+2 \alpha\right)= \\
& =1+\sin 2 \alpha+\cos 2 \alpha=\cos ^{2} \alpha+\sin ^{2} \alpha+2 \sin \alpha \cos \alpha+\cos ^{2} \alpha-\sin ^{2} \alpha= \\
& =\left(\cos ^{2} \alpha+2 \sin \alpha \cos \alpha+\sin ^{2} \alpha\right)+(\cos \alpha-\sin \alpha)(\cos \alpha+\sin \alpha)= \\
& =(\cos \alpha+\sin \alpha)^{2}+(\cos \alpha-\sin \alpha)(\cos \alpha+\sin \alpha)=
\end{aligned}
$$

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

$=2(\cos \alpha+\sin \alpha) \cos \alpha=2\left(\cos \alpha+\cos \left(90^{\circ}-\alpha\right)\right) \cos \alpha=$

$=2 \cdot 2 \cos 45^{\circ} \cos \left(\alpha-45^{\circ}\right) \cos \alpha=4 \cdot \frac{\sqrt{2}}{2} \cos \left(\alpha-45^{\circ}\right) \cos \alpha=$

$=2 \sqrt{2} \cos \alpha \cos \left(\alpha-45^{\circ}\right)=2 \sqrt{2} \cos \alpha \cos \left(45^{\circ}-\alpha\right)$.

Answer: $2 \sqrt{2} \cos \alpha \cos \left(45^{\circ}-\alpha\right)$.