3.126. $1+\cos \left(\frac{\pi}{2}+3 \alpha\right)-\sin \left(\frac{3}{2} \pi-3 \alpha\right)+\operatorname{ctg}\left(\frac{5}{2} \pi+3 \alpha\right)$.
## Solution.

$$
1+\cos \left(\frac{\pi}{2}+3 \alpha\right)-\sin \left(\frac{3}{2} \pi-3 \alpha\right)+\operatorname{ctg}\left(\frac{5}{2} \pi+3 \alpha\right)=
$$

$$
\begin{aligned}
& =1-\sin 3 \alpha+\cos 3 \alpha-\operatorname{tg} 3 \alpha=1-\sin 3 \alpha+\cos 3 \alpha-\frac{\sin 3 \alpha}{\cos 3 \alpha}= \\
& =\frac{(\cos 3 \alpha-\sin 3 \alpha) \cos 3 \alpha+\cos 3 \alpha-\sin 3 \alpha}{\cos 3 \alpha}=\frac{(\cos 3 \alpha-\sin 3 \alpha)(\cos 3 \alpha+1)}{\cos 3 \alpha}= \\
& =\frac{\left(\cos 3 \alpha-\cos \left(\frac{\pi}{2}-3 \alpha\right)\right)(\cos 3 \alpha+\cos 2 \pi)}{\cos 3 \alpha}= \\
& =\frac{-2 \sin \frac{\pi}{4} \sin \left(3 \alpha-\frac{\pi}{4}\right) \cdot 2 \cos \left(\pi+\frac{3 \alpha}{2}\right) \cos \left(\pi-\frac{3 \alpha}{2}\right)}{\cos 3 \alpha}= \\
& =\frac{-2 \cdot \frac{\sqrt{2}}{2} \cdot \sin \left(3 \alpha-\frac{\pi}{4}\right) \cdot 2 \cdot\left(-\cos \frac{3 \alpha}{2}\right)\left(-\cos \frac{3 \alpha}{2}\right)}{\cos 3 \alpha}= \\
& =\frac{2 \sqrt{2} \cos ^{2} \frac{3 \alpha}{2} \sin \left(\frac{\pi}{4}-3 \alpha\right)}{\cos 3 \alpha} .
\end{aligned}
$$

Answer: $\frac{2 \sqrt{2} \cos ^{2} \frac{3 \alpha}{2} \sin \left(\frac{\pi}{4}-3 \alpha\right)}{\cos 3 \alpha}$.