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[函数] 如何化简一个三角函数表达式

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TSC999 posted 2025-3-3 08:56 |Read mode
Last edited by hbghlyj 2025-3-3 18:29下面这个表达式能化简成一个由数字组成的根式表达式。其值为 \(1.47603491931....\)
\[
-(2+\operatorname{Cos}[2 \operatorname{ArcTan}[\sqrt{6+\sqrt{33}}]])^2 \operatorname{Csc}[4 \operatorname{ArcTan}[\sqrt{6+\sqrt{33}}]]
\]其中 \(\rm ArcTan\) 是反正切函数。\(\rm Csc\) 是正弦函数的倒数。
最好是能用 mathematica 的指令化简。

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战巡 posted 2025-3-3 11:42
p0220.png

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kuing posted 2025-3-3 14:21
人工化简也不难啊,因为
\[\cos(\arctan x)=\frac1{\sqrt{1+x^2}},\]

\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*}

\[\text{原式}=-\frac{(2+\cos2\theta)^2}{\sin4\theta}=\frac{(1+2p^2)^2}{4pq(2p^2-1)},\]
代入上面的式子化简得
\[\text{原式}=\frac{\bigl(9+\sqrt{33}\bigr)^2}{4\bigl(5+\sqrt{33}\bigr)\sqrt{6+\sqrt{33}}}=\sqrt{\frac38\bigl(69-11\sqrt{33}\bigr)}.\]

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