Example 3.2 Find the sum $\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k} \sin (k x+a)$, where $a$ is a constant.
Let $f(k)=\sin (k x+a)$, then
$$
\begin{aligned}
\sum_{k=0}^{n} & (-1)^{n-k}\binom{n}{k} \sin (k x+a) \\
& =\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k} f(k) \\
& =\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k} E^{k} f(0)=(E-I)^{n} f(0) \\
& =\Delta^{n} f(0) \\
\Delta f(k) & =f(k+1)-f(k) \\
& =\sin [(k+1) x+a]-\sin (k x+a) \\
& =2 \sin \frac{x}{2} \cos \left(k x+\frac{x}{2}+a\right) \\
& =2 \sin \frac{x}{2} \sin \left(k x+a+\frac{x+\pi}{2}\right) .
\end{aligned}
$$

Let $\Delta^{s} f(k)=2^{s} \sin ^{s} \frac{x}{2} \sin \left[k x+a+\frac{s(x+\pi)}{2}\right]$, then
$$
\begin{aligned}
\Delta^{s+1} f(k)= & \Delta\left(\Delta^{s} f(k)\right)=\Delta^{s} f(k+1)-\Delta^{s} f(k) \\
= & 2^{s} \sin ^{s} \frac{x}{2} \sin \left[(k+1) x+a+\frac{s(x+\pi)}{2}\right]- \\
& 2^{s} \sin ^{s} \frac{x}{2} \sin \left[k x+a+\frac{s(x+\pi)}{2}\right] \\
= & 2^{s} \sin ^{s} \frac{x}{2} \cdot 2 \sin \frac{x}{2} \cos \left[k x+\frac{x}{2}+a+\frac{s(x+\pi)}{2}\right] \\
= & 2^{s+1} \sin ^{s+1} \frac{x}{2} \sin \left[k x+a+\frac{(s+1)(x+\pi)}{2}\right]
\end{aligned}
$$

Thus, by mathematical induction,
$$
\Delta^{n} f(k)=2^{n} \sin ^{n} \frac{x}{2} \sin \left[k x+a+\frac{n(x+\pi)}{2}\right] .
$$

Therefore,
$$
\Delta^{n} f(0)=2^{n} \sin ^{n} \frac{x}{2} \sin \left[a+\frac{n(x+\pi)}{2}\right],
$$

which means
$$
\begin{array}{l}
\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k} \sin (k x+a) \\
\quad=2^{n} \sin ^{n} \frac{x}{2} \sin \left[a+\frac{n(x+\pi)}{2}\right] .
\end{array}
$$