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Question Number 164716 by cortano1 last updated on 21/Jan/22

$$findminimumvalueof$$$$f\left(x\right)=4\sin 2x-5\sin x-5\cos x+6$$

Answered by bobhans last updated on 21/Jan/22

$$\mathrm{f}\left(\mathrm{x}\right)=4\cos (\frac{\pi }{2}-2\mathrm{x})-5\sqrt{2}\cos (\frac{\pi }{4}-\mathrm{x})+6$$$$\mathrm{f}\left(\mathrm{x}\right)=4\cos 2(\frac{\pi }{4}-\mathrm{x})-5\sqrt{2}\cos (\frac{\pi }{4}-\mathrm{x})+6$$$$\frac{\pi }{4}-\mathrm{x}=\phi \Rightarrow \mathrm{f}\left(\phi \right)=4\cos 2\phi -5\sqrt{2}\cos \phi +6$$$$\mathrm{f}\left(\phi \right)=8\cos {}^2\phi -5\sqrt{2}\cos \phi +2$$$$\mathrm{f}\left(\phi \right)\mathrm{will}\mathrm{be}\min \mathrm{when}\cos \phi =\frac{5\sqrt{2}}{16}$$$$\min \mathrm{f}\left(\phi \right)=8{\left(\frac{5\sqrt{2}}{16}\right)}^2-5\sqrt{2}\left(\frac{5\sqrt{2}}{16}\right)+2=\frac{7}{16}$$$$$$

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