问题及解答

用 $\cos x$ 表示 $\cos 2x$, $\cos 4x$, $\cos 6x$.

\[
\cos 2x=2\cos^2 x-1,
\]

\[
\cos 4x=8\cos^4 x-8\cos^2 x+1,
\]

\[
\cos 6x=32\cos^6 x-48\cos^4 x+18\cos^2 x-1.
\]

\[
\begin{split}
\cos(4x)&=2\cos^2(2x)-1\\
&=2\bigl(2\cos^2 x-1\bigr)^2-1\\
&=2\bigl(4\cos^4 x-4\cos^2 x+1\bigr)-1\\
&=8\cos^4 x-8\cos^2 x+1.
\end{split}
\]

\[
\begin{split}
\cos 6x&=\cos(4x+2x)=\cos 4x\cdot\cos 2x-\sin 4x\cdot\sin 2x\\
&=(8\cos^4 x-8\cos^2 x+1)(2\cos^2 x-1)-2\sin^2(2x)\cdot\cos 2x\\
&=(16\cos^6 x-8\cos^4 x-16\cos^4 x+8\cos^2 x+2\cos^2 x-1)-2(1-\cos^2(2x))\cdot\cos 2x\\
&=16\cos^6 x-24\cos^4 x+10\cos^2 x-1-2\cos 2x+2\cos^3(2x)\\
&=16\cos^6 x-24\cos^4 x+10\cos^2 x-1-2(2\cos^2 x-1)+2\bigl(2\cos^2 x-1\bigr)^3\\
&=16\cos^6 x-24\cos^4 x+10\cos^2 x-1-4\cos^2 x+2+2\bigl(8\cos^6 x-3\cdot 4\cos^4 x+3\cdot 2\cos^2 x-1\bigr)\\
&=16\cos^6 x-24\cos^4 x+6\cos^2 x+1+16\cos^6 x-24\cos^4 x+12\cos^2 x-2\\
&=32\cos^6 x-48\cos^4 x+18\cos^2 x-1.
\end{split}
\]