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[不等式] 三角函数最值

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yuzi Post time 2015-8-24 21:36 |Read mode
QQ截图20150824213438.jpg

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kuing Post time 2015-8-25 01:47
根据三角恒等式
\[\sin A+\sin B+\sin C=4\cos \frac A2\cos \frac B2\cos \frac C2,\]

\begin{align*}
\text{原式} & =\frac14\sum\tan \frac A2\tan \frac B2\sec \frac C2 \\
& =\frac14\sum\tan \frac A2\tan \frac B2\sqrt{\tan ^2\frac C2+1} \\
& \leqslant \frac14\sqrt{\sum\tan \frac A2\tan \frac B2\sum\tan \frac A2\tan \frac B2\left( \tan ^2\frac C2+1 \right)} \\
& =\frac14\sqrt{\tan \frac A2\tan \frac B2\tan \frac C2\left( \tan \frac A2+\tan \frac B2+\tan \frac C2 \right)+1} \\
& \leqslant \frac14\sqrt{\frac13\left( \sum\tan \frac A2\tan \frac B2 \right)^2+1} \\
& =\frac{\sqrt3}6,
\end{align*}
当正三角形时取等。

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 Author| yuzi Post time 2015-8-25 21:30
不等式渣。。。。好难

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其妙 Post time 2015-8-25 23:46
v6不是写了一个接发?

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