10. Simplify: (1) $\sin 5 A-5 \sin 3 A+10 \sin A$;
(2) $1+C_{n}^{1} \cos x+C_{n}^{2} \cos 2 x+C_{n}^{3} \cos 3 x+\cdots+C_{n}^{n} \cos n x \quad\left(n \in \mathbf{N}^{*}\right)$.
10. (1) Let $z=\cos A+\mathrm{i} \sin A$, then $\bar{z}=\cos A-\mathrm{i} \sin A, \sin A=\frac{z-\bar{z}}{2 \mathrm{i}}, \sin 3 A=\frac{z^{3}-\overline{z^{3}}}{2 \mathrm{i}}, \sin 5 A=\frac{z^{5}-\overline{z^{5}}}{2 \mathrm{i}}$
$$
\begin{array}{l}
\text { Hence } \sin 5 A-5 \sin 3 A+10 \sin A=\frac{z^{5}-\overline{z^{5}}}{2 \mathrm{i}}-5 \cdot \frac{z^{3}-\overline{z^{3}}}{2 \mathrm{i}}+10 \cdot \frac{z-\bar{z}}{2 \mathrm{i}}=\frac{1}{2 \mathrm{i}}\left(z^{5}-5 z^{3}+10 z-10 \bar{z}+\right. \\
\left.5 \overline{z^{3}}-\overline{z^{5}}\right)=\frac{1}{2 \mathrm{i}}(z-\bar{z})^{5}=16 \cdot\left(\frac{z-\bar{z}}{2 \mathrm{i}}\right)^{5}=16 \sin ^{5} A
\end{array}
$$
(2) Let $z=\cos x+i \sin x$, then on one hand, by the binomial theorem we have:
$$
\begin{array}{l}
\quad(1+z)^{n}=1+C_{n}^{1} z+C_{n}^{2} z^{2}+\cdots+C_{n}^{k} z^{k}+\cdots+C_{n}^{n} z^{n}=\left(1+C_{n}^{1} \cos x+C_{n}^{2} \cos 2 x+\cdots+C_{n}^{n} \cos n x\right)+ \\
\mathrm{i}\left(C_{n}^{1} \sin x+C_{n}^{2} \sin 2 x+\cdots+C_{n}^{n} \sin n x\right) .
\end{array}
$$

On the other hand, by De Moivre's theorem, we have $(1+z)^{n}=(1+\cos x+\mathrm{i} \sin x)^{n}=\left(2 \cos ^{2} \frac{x}{2}+2 \mathrm{isin} \frac{x}{2} \cos \frac{x}{2}\right)^{n}=$
$$
2^{n} \cos ^{n} \frac{x}{2}\left(\cos \frac{x}{2}+\mathrm{i} \sin \frac{x}{2}\right)^{n}=2^{n} \cos ^{n} \frac{x}{2}\left(\cos \frac{n x}{2}+\mathrm{i} \sin \frac{n x}{2}\right)
$$

Therefore, $\quad 1+C_{n}^{1} \cos x+C_{n}^{2} \cos 2 x+\cdots+C_{n}^{n} \cos n x=2^{n} \cos ^{n} \frac{x}{2} \cdot \cos \frac{n x}{2}, n \in \mathbf{N}^{\cdot}$