3.128. $1-\cos \left(2 \alpha-270^{\circ}\right)+\sin \left(2 \alpha+270^{\circ}\right)$.
Solution.

$1-\cos \left(2 \alpha-270^{\circ}\right)+\sin \left(2 \alpha+270^{\circ}\right)=1-\cos \left(270^{\circ}-2 \alpha\right)+\sin \left(270^{\circ}+2 \alpha\right)=$ $=1+\sin 2 \alpha-\cos 2 \alpha=\sin ^{2} \alpha+\cos ^{2} \alpha+2 \sin \alpha \cos \alpha-\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right)=$ $=(\sin \alpha+\cos \alpha)^{2}+\sin ^{2} \alpha-\cos ^{2} \alpha=(\sin \alpha+\cos \alpha)^{2}+$ $+(\sin \alpha+\cos \alpha)(\sin \alpha-\cos \alpha)=(\sin \alpha+\cos \alpha)(\sin \alpha+\cos \alpha+\sin \alpha-\cos \alpha)=$ $=2(\sin \alpha+\cos \alpha) \sin \alpha=2\left(\sin \alpha+\sin \left(90^{\circ}-\alpha\right)\right) \sin \alpha=$

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

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

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