|
人工化简也不难啊,因为
\[\cos(\arctan x)=\frac1{\sqrt{1+x^2}},\]
记
\begin{align*}
\theta&=\arctan\sqrt{6+\sqrt{33}},\\
p&=\cos\theta=\frac1{\sqrt{7+\sqrt{33}}},\\
q&=\sin\theta=\sqrt{\frac{6+\sqrt{33}}{7+\sqrt{33}}},
\end{align*}
则
\[\text{原式}=-\frac{(2+\cos2\theta)^2}{\sin4\theta}=\frac{(1+2p^2)^2}{4pq(2p^2-1)},\]
代入上面的式子化简得
\[\text{原式}=\frac{\bigl(9+\sqrt{33}\bigr)^2}{4\bigl(5+\sqrt{33}\bigr)\sqrt{6+\sqrt{33}}}=\sqrt{\frac38\bigl(69-11\sqrt{33}\bigr)}.\] |
|