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[函数] 反三角函数$y=\arccos(\sin A)+\arccos(\sin B)+\arccos(\sin C)$

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isee Posted 2019-8-31 15:50 |Read mode
在$\triangle ABC$中,$y=\arccos(\sin A)+\arccos(\sin B)+\arccos(\sin C)$,求$y$的范围。

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看欧拉和的解法,发现有些解法需要反三角函数。

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kuing Posted 2019-8-31 16:12
因为 `\pi/2-A\in(-\pi/2,\pi/2)` 所以
\[\arccos(\sin A)=\arccos\left( \cos\left( \frac\pi2-A \right) \right)=\left| \frac\pi2-A \right|,\]其余同理,下略。

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