问题及解答

证明 $\cos 3\theta=4\cos^3\theta-3\cos\theta$.

至少有两种证法.

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

应用 De Moivre 定理,

\[
(\cos\theta+i\sin\theta)^3=\cos 3\theta+i\sin 3\theta
\]

另一方面

\[
\begin{split}
(\cos\theta+i\sin\theta)^3&=\cos^3\theta+3\cos^2\theta\cdot i\sin\theta+3\cos\theta\cdot(i\sin\theta)^2+(i\sin\theta)^3\\
&=\cos^3\theta+i3\cos^2\theta\sin\theta-3\cos\theta\sin^2\theta-i\sin^3\theta\\
&=(\cos^3\theta-3\cos\theta\sin^2\theta)+i(3\cos^2\theta\sin\theta-\sin^3\theta)\\
&=\bigl[\cos^3\theta-3\cos\theta\cdot(1-\cos^2\theta)\bigr]+i\bigl[\sin\theta\cdot(3\cos^2\theta-\sin^2\theta)\bigr]\\
&=\bigl[4\cos^3\theta-3\cos\theta\bigr]+i\bigl[\sin\theta\cdot(3-4\sin^2\theta)\bigr]\\
&=\bigl[4\cos^3\theta-3\cos\theta\bigr]+i\bigl[3\sin\theta-4\sin^3\theta\bigr]\\
\end{split}
\]

比较实部和虚部, 得

\[
\begin{aligned}
\cos 3\theta&=4\cos^3\theta-3\cos\theta\\
\sin 3\theta&=3\sin\theta-4\sin^3\theta
\end{aligned}
\]