請教三角極值問題sin(x/2)(1+cosx),0<x<180，求最大值
請問如果不用微分要解這題三角極值該如何下手呢?原本想用算幾，卻又覺得怪，產三下列三種想法
還是都是錯誤的方向呢?
\(sin(\frac{x}{2})(1+cosx)=2sin(\frac{x}{2})cos^{2}(\frac{x}{2})\)
\(\begin{cases}
\frac{\displaystyle 2sin(\frac{x}{2})+cos^2(\frac{x}{2})}{\displaystyle 2}\geq \sqrt{2sin(\frac{x}{2})cos^{2}(\frac{x}{2})}......plan~A\\
\frac{\displaystyle 2sin(\frac{x}{2})+cos(\frac{x}{2})+cos(\frac{x}{2})}{\displaystyle 3}\geq \sqrt[3]{2sin(\frac{x}{2})cos^{2}(\frac{x}{2})}......plan~B\\
\frac{\displaystyle sin(\frac{x}{2})+\sqrt{2}cos(\frac{x}{2})+\sqrt{2}cos(\frac{x}{2})}{\displaystyle 3} \geq \sqrt[3]{2sin(\frac{x}{2})cos^{2}(\frac{x}{2})}.....plan~C \end{cases}\)
[[i] 本帖最後由 deca0206 於 2015-10-2 05:36 PM 編輯 [/i]]
回復 1# deca0206 的帖子
考慮\(\sin \frac{x}{2}=\sqrt{\frac{1-\cos x}{2}}\)回復 2# thepiano 的帖子
哇~驚醒夢中人，感謝老師回復 1# deca0206 的帖子
也可以把 plan B 改成\(\frac{2{{\sin }^{2}}\left( \frac{x}{2} \right)+{{\cos }^{2}}\left( \frac{x}{2} \right)+{{\cos }^{2}}\left( \frac{x}{2} \right)}{3}\ge \sqrt[3]{2{{\sin }^{2}}\left( \frac{x}{2} \right){{\cos }^{2}}\left( \frac{x}{2} \right){{\cos }^{2}}\left( \frac{x}{2} \right)}\)回復 4# thepiano 的帖子
好主意頁:
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