50 (1178). Simplify the expression

$$
\sin 2 \alpha+\sin 4 \alpha+\ldots+\sin 2 n \alpha
$$
Solution. Multiply and divide the given expression by $\sin \alpha$ and use the formula for transforming the product of sines. We will have:

$$
\begin{gathered}
\sin 2 \alpha+\sin 4 \alpha+\ldots+\sin 2 n \alpha= \\
=\frac{\sin \alpha(\sin 2 \alpha+\sin 4 \alpha+\sin 6 \alpha+\ldots+\sin (2 n \alpha))}{\sin \alpha}= \\
=\frac{\sin \alpha \sin 2 \alpha+\sin \alpha \sin 4 \alpha+\sin \alpha \sin 6 \alpha+\ldots+\sin \alpha \sin (2 n \alpha)}{\sin \alpha}= \\
=\frac{1}{2 \sin \alpha}((\cos \alpha-\cos 3 \alpha)+(\cos 3 \alpha-\cos 5 \alpha)+ \\
+(\cos 5 \alpha-\cos 7 \alpha)+\ldots+(\cos ((2 n-1) \alpha)-\cos ((2 n+1) \alpha))= \\
=\frac{1}{2 \sin \alpha}(\cos \alpha-\cos ((2 n+1) \alpha))= \\
=\frac{1}{2 \sin \alpha} \cdot 2 \sin ((n+1) \alpha) \sin (n \alpha)=\frac{\sin ((n+1) \alpha) \sin (n \alpha)}{\sin \alpha} .
\end{gathered}
$$

Answer: $\frac{\sin ((n+1) \alpha) \sin (n \alpha)}{\sin \alpha}$.