问题及解答

当 $\theta\neq 2k\pi$, $k\in\mathbb{Z}$ 时, 证明下面的恒等式,

\[
\sin x+\sin(x+\theta)+\sin(x+2\theta)+\cdots+\sin(x+(n-1)\theta)=\frac{\sin\frac{n\theta}{2}\cdot\sin(x+\frac{n-1}{2}\theta)}{\sin\frac{\theta}{2}}.
\]

[分析]  恒等式左边是关于等差数列 $\{x+k\theta\mid k=0,1,2,\ldots,n-1\}$ 的正弦值的和.

回顾三角函数的积化和差公式

\[
\sin x\cdot\sin y=-\frac{1}{2}\bigl[\cos(x+y)-\cos(x-y)\bigr]=\frac{1}{2}\bigl[\cos(x-y)-\cos(x+y)\bigr],\tag{*}
\]

这可由和差化积公式

\[
\cos\alpha-\cos\beta=-2\sin\frac{\alpha+\beta}{2}\cdot\sin\frac{\alpha-\beta}{2}
\]

推出.

注意到 $(*)$ 式右边是两个式子的差, 故如果是很多项都形如这种形式, 可以构造成前后抵消的形式.

我们假设 $\sin x+\sin(x+\theta)+\sin(x+2\theta)+\cdots+\sin(x+(n-1)\theta)$ 乘以 $\sin t$ 后可以前后抵消,

\[
\begin{aligned}
\sin t\cdot\sin x&=\frac{1}{2}\bigl[\cos(t-x)-\cos(t+x)\bigr]=\frac{1}{2}\bigl[\cos(x-t)-\cos(t+x)\bigr]\\
\sin t\cdot\sin(x+\theta)&=\frac{1}{2}\bigl[\cos(x+\theta-t)-\cos(t+x+\theta)\bigr]\\
\sin t\cdot\sin(x+2\theta)&=\frac{1}{2}\bigl[\cos(x+2\theta-t)-\cos(t+x+2\theta)\bigr]\\
&\vdots\\
\sin t\cdot\sin(x+(n-1)\theta)&=\frac{1}{2}\bigl[\cos(x+(n-1)\theta-t)-\cos(t+x+(n-1)\theta)\bigr]\\
\end{aligned}
\]

要使得前后抵消, 只需令 $x+k\theta-t=t+x+(k-1)\theta$, 这推出 $t=\frac{1}{2}\theta$.

因此, 证明方法是乘以 $\sin\frac{1}{2}\theta$.

证明: 在欲证恒等式左边乘以 $\sin\frac{1}{2}\theta$, 得

\[
\begin{split}
&\sin(\frac{1}{2}\theta)\cdot\sum_{k=0}^{n-1}\sin(x+k\theta)\\
=&\sum_{k=0}^{n-1}\sin\frac{1}{2}\theta\cdot\sin(x+k\theta)\\
=&\sum_{k=0}^{n-1}\frac{1}{2}\frac{1}{2}\Bigl[\cos\bigl(x+(k-\frac{1}{2})\theta\bigr)-\cos\bigl(x+(k+\frac{1}{2})\theta\bigr)\Bigr]\\
=&\frac{1}{2}\Bigl(\bigl[\cos(x-\frac{1}{2}\theta)-\cos(x+\frac{1}{2}\theta)\bigr]+\bigl[\cos(x+\frac{1}{2}\theta)-\cos(x+\frac{3}{2}\theta)\bigr]+\cdots+\bigl[\cos(x+(n-\frac{3}{2})\theta)-\cos(x+(n-\frac{1}{2})\theta)\bigr]\Bigr)\\
=&\frac{1}{2}\bigl[\cos(x-\frac{1}{2}\theta)-\cos(x+(n-\frac{1}{2})\theta)\bigr]\\
=&\frac{1}{2}\cdot(-2)\sin(x+\frac{n-1}{2}\theta)\sin(-\frac{n}{2}\theta)\\
=&\sin(\frac{n}{2}\theta)\cdot\sin(x+\frac{n-1}{2}\theta).
\end{split}
\]

因此

\[
\sum_{k=0}^{n-1}\sin(x+k\theta)=\frac{\sin(\frac{n}{2}\theta)\cdot\sin(x+\frac{n-1}{2}\theta)}{\sin(\frac{1}{2}\theta)}.
\]