15. Simplify: $C_{n}^{0} \sin \alpha-C_{n}^{1} \sin (\alpha+\theta)+\cdots+(-1)^{n} C_{n}^{n} \sin (\alpha+n \theta)$.
15. Let the complex number $z=C_{n}^{0}(\cos \alpha+$
$$
i \sin \alpha)-C_{n}^{1}[\cos (\alpha+\theta)+i \sin (\alpha+
$$
$\theta)]+C_{n}^{2}[\cos (\alpha+2 \theta)+i \sin (\alpha+2 \theta)]-\cdots+(-1)^{n} C_{n}^{n}[\cos (\alpha+n \theta)+i \sin (\alpha+$ $n \theta)]$,
then $z=(\cos \alpha+i \sin \alpha)[1-(\cos \theta+i \sin \theta)]^{n}$
$$
=2^{n} \sin ^{n} \frac{\theta}{2}\left[\cos \left(\frac{3 n \pi}{2}+\frac{n \theta}{2}+\alpha\right)+i \sin \left(\frac{3 n \pi}{2}+\frac{n \theta}{2}+\alpha\right)\right] \text {, }
$$

Comparing the imaginary part, we get the original expression $=2^{n} \sin ^{n} \frac{\theta}{2} \sin \left(\frac{3 n \pi}{2}+\frac{n \theta}{2}+\alpha\right)$.