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Question Number 144452 by imjagoll last updated on 25/Jun/21

$$\mathrm{Find}\mathrm{minimum}\mathrm{value}\mathrm{of}$$$$\mathrm{f}\left(\mathrm{x}\right)=\sin (\mathrm{x}+3)-\sin (\mathrm{x}+1)-2\cos (\mathrm{x}+2)$$$$\mathrm{where}\mathrm{x}ϵ\mathrm{R}$$

Answered by EDWIN88 last updated on 25/Jun/21

$$\mathrm{f}\left(\mathrm{x}\right)=\sin (\mathrm{x}+3)-\sin (\mathrm{x}+1)-2\cos (\mathrm{x}+2)$$$$\mathrm{f}\left(\mathrm{x}\right)=2\cos (\mathrm{x}+2)\sin \left(1\right)-2\cos (\mathrm{x}+2)$$$$\mathrm{f}\left(\mathrm{x}\right)=2(\sin \left(1\right)-1)\cos (\mathrm{x}+2)$$$$\mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{l}\min =2\sin \left(1\right)-2\\ \max =-2\sin \left(1\right)+2\end{array}$$

Answered by Olaf_Thorendsen last updated on 25/Jun/21

$$f\left(x\right)=\sin (x+3)-\sin (x+1)-2\cos (x+2)$$$$f\left(x\right)=2\sin \left(\frac{(x+3)-(x+1)}{2}\right)\cos \left(\frac{(x+3)+(x+1)}{2}\right)$$$$-2\cos (x+2)$$$$f\left(x\right)=2\sin \left(1\right)\cos (x+2)-2\cos (x+2)$$$$f\left(x\right)=2(\sin \left(1\right)-1)\cos (x+2)$$$$$$$$\mathrm{Min}\left(f\right)=2(\sin \left(1\right)-1)$$$$\mathrm{when}\cos (x+2)=1$$$$x+2=\frac{\pi }{2}+2k\pi ,k\in \mathbb{Z}$$$$x=\frac{\pi }{2}+2(k\pi -1),k\in \mathbb{Z}$$

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