3.380. It is known that $\operatorname{tg} \alpha=\frac{p}{q}$. Find $\sin 2 \alpha, \cos 2 \alpha$ and $\operatorname{tg} 2 \alpha$.
## Solution.

By the formulas of universal trigonometric substitution we have

$$
\begin{aligned}
& \sin 2 \alpha=\frac{2 \tan \alpha}{1+\tan^{2} \alpha}, \alpha \neq(2 n+1) \pi ; \cos 2 \alpha=\frac{1-\tan^{2} \alpha}{1+\tan^{2} \alpha}, \alpha \neq(2 n+1) \pi ; \\
& \tan 2 \alpha=\frac{2 \tan \alpha}{1-\tan^{2} \alpha}, \alpha \neq \frac{\pi}{4}+\frac{\pi}{2} k, k \in Z, \alpha \neq \frac{\pi}{2}+\pi n, n \in Z
\end{aligned}
$$

Given that $\tan \alpha=\frac{p}{q}$, therefore $\sin 2 \alpha=\frac{\frac{2 p}{q}}{1+\frac{p^{2}}{q^{2}}}=\frac{2 p q}{p^{2}+q^{2}}$;

$\cos 2 \alpha=\frac{1-\frac{p^{2}}{q^{2}}}{1+\frac{p^{2}}{q^{2}}}=\frac{q^{2}-p^{2}}{q^{2}+p^{2}} ; \tan 2 \alpha=\frac{\frac{2 p}{q}}{1-\frac{p^{2}}{q^{2}}}=\frac{2 p q}{q^{2}-p^{2}}$.

Answer: $\sin 2 \alpha=\frac{2 p q}{p^{2}+q^{2}} ; \cos 2 \alpha=\frac{q^{2}-p^{2}}{q^{2}+p^{2}} ; \tan 2 \alpha=\frac{2 p q}{q^{2}-p^{2}}$.