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Question Number 162721 by mnjuly1970 last updated on 31/Dec/21

$$$$$$calculate$$$$f\left(x\right)=\frac{1}{4(1+cos\left(\frac{x}{2}\right))}+\frac{1}{9(1-cos\left(\frac{x}{2}\right))}(x\ne 2k\pi ,k\in \mathbb{Z})$$$$f_{min}=?$$$$\mathcal{A}dapted\mathcal{F}rom\mathcal{I}nstagram$$$$$$

Answered by Ar Brandon last updated on 31/Dec/21

$$f\left(x\right)=\frac{1}{4(1+\cos \frac{x}{2})}+\frac{1}{9(1-\cos \frac{x}{2})}$$$$=\frac{1}{{8\cos }^2\frac{x}{4}}+\frac{1}{{18\sin }^2\frac{x}{4}}=\frac{8+{10\sin }^2\frac{x}{4}}{{36\sin }^2\frac{x}{2}}$$$$\mathrm{g}\left(x\right)={36\sin }^2\frac{x}{2}\Rightarrow {\mathrm{g}}^{\prime }\left(x\right)=18\sin x=0\Rightarrow x=\pi$$$$f\left(x\right)\min \Rightarrow \mathrm{g}\left(x\right)\mathrm{ma}x\Rightarrow x=\pi$$$$f_{min}=\frac{8+{10\sin }^2\frac{\pi }{4}}{{36\sin }^2\frac{\pi }{2}}=\frac{8+5}{36}=\frac{13}{36}$$

Answered by mr W last updated on 31/Dec/21

$$f\left(x\right)=\frac{1}{4\times 2{\cos }^2\frac{x}{4}}+\frac{1}{9\times 2{\sin }^2\frac{x}{4}}$$$$f\left(x\right)=\frac{1}{8}(\frac{1}{1-{\sin }^2\frac{x}{4}}+\frac{4}{9{\sin }^2\frac{x}{4}})$$$$f\left(t\right)=\frac{1}{8}(\frac{1}{1-t}+\frac{4}{9t})witht={\sin }^2\frac{x}{4}\in (0,1)$$$$f^{\prime }\left(t\right)=0\Rightarrow \frac{1}{{(1-t)}^2}-\frac{4}{9t^2}=0$$$$5t^2+8t-4=0$$$$(5t-2)(t+2)=0$$$$\Rightarrow t=\frac{2}{5},-2\left(rejected\right)$$$$f_{min}=\frac{1}{8}(\frac{1}{1-\frac{2}{5}}+\frac{4}{9\times \frac{2}{5}})=\frac{25}{72}✓$$

Commented by mnjuly1970 last updated on 31/Dec/21

$$bravosirWnicesolution$$$$asalways...grateful...$$

Answered by mnjuly1970 last updated on 31/Dec/21

$$f\left(x\right)=\frac{1}{8}+\frac{1}{8}{tan}^2\left(\frac{x}{4}\right)+\frac{1}{18}+\frac{1}{18}{cot}^2\left(\frac{x}{4}\right)$$$$⩾\frac{13}{72}+2\sqrt{\frac{1}{8}.\frac{1}{18}}=\frac{13}{72}+\frac{1}{6}$$$$f_{min}=\frac{25}{72}$$

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