问题及解答

证明三倍角公式:

\[
\sin 3\theta=3\sin\theta-4\sin^3\theta.
\]

Pf.

\[
\begin{split}
\sin 3\theta&=\sin(\theta+2\theta)\\
&=\sin\theta\cos(2\theta)+\cos\theta\sin(2\theta)\\
&=\sin\theta\cdot(1-2\sin^2\theta)+\cos\theta\cdot 2\sin\theta\cos\theta\\
&=\sin\theta-2\sin^3\theta+2\sin\theta\cdot(1-\sin^2\theta)\\
&=3\sin\theta-4\sin^3\theta.
\end{split}
\]