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Question Number 144452 by imjagoll last updated on 25/Jun/21
Find minimum value of f(x)=sin (x+3)−sin (x+1)−2cos (x+2) where xεR
$$\:\mathrm{Find}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sin}\:\left(\mathrm{x}+\mathrm{3}\right)−\mathrm{sin}\:\left(\mathrm{x}+\mathrm{1}\right)−\mathrm{2cos}\:\left(\mathrm{x}+\mathrm{2}\right) \\ $$$$\mathrm{where}\:\mathrm{x}\epsilon\mathrm{R} \\ $$
Answered by EDWIN88 last updated on 25/Jun/21
f(x)=sin (x+3)−sin (x+1)−2cos (x+2) f(x)=2cos (x+2)sin (1)−2cos (x+2) f(x)=2(sin (1)−1)cos (x+2) f(x)= { ((min=2sin (1)−2)),((max=−2sin (1)+2)) :}
$$\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sin}\:\left(\mathrm{x}+\mathrm{3}\right)−\mathrm{sin}\:\left(\mathrm{x}+\mathrm{1}\right)−\mathrm{2cos}\:\left(\mathrm{x}+\mathrm{2}\right) \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2cos}\:\left(\mathrm{x}+\mathrm{2}\right)\mathrm{sin}\:\left(\mathrm{1}\right)−\mathrm{2cos}\:\left(\mathrm{x}+\mathrm{2}\right) \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2}\left(\mathrm{sin}\:\left(\mathrm{1}\right)−\mathrm{1}\right)\mathrm{cos}\:\left(\mathrm{x}+\mathrm{2}\right) \\ $$$$\:\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{\mathrm{min}=\mathrm{2sin}\:\left(\mathrm{1}\right)−\mathrm{2}}\\{\mathrm{max}=−\mathrm{2sin}\:\left(\mathrm{1}\right)+\mathrm{2}}\end{cases} \\ $$
Answered by Olaf_Thorendsen last updated on 25/Jun/21
f(x) = sin(x+3)−sin(x+1)−2cos(x+2) f(x) = 2sin((((x+3)−(x+1))/2))cos((((x+3)+(x+1))/2)) −2cos(x+2) f(x) = 2sin(1)cos(x+2)−2cos(x+2) f(x) = 2(sin(1)−1)cos(x+2) Min(f) = 2(sin(1)−1) when cos(x+2) = 1 x+2 = (π/2)+2kπ, k∈Z x = (π/2)+2(kπ−1), k∈Z
$${f}\left({x}\right)\:=\:\mathrm{sin}\left({x}+\mathrm{3}\right)−\mathrm{sin}\left({x}+\mathrm{1}\right)−\mathrm{2cos}\left({x}+\mathrm{2}\right) \\ $$$${f}\left({x}\right)\:=\:\mathrm{2sin}\left(\frac{\left({x}+\mathrm{3}\right)−\left({x}+\mathrm{1}\right)}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\left({x}+\mathrm{3}\right)+\left({x}+\mathrm{1}\right)}{\mathrm{2}}\right) \\ $$$$−\mathrm{2cos}\left({x}+\mathrm{2}\right) \\ $$$${f}\left({x}\right)\:=\:\mathrm{2sin}\left(\mathrm{1}\right)\mathrm{cos}\left({x}+\mathrm{2}\right)−\mathrm{2cos}\left({x}+\mathrm{2}\right) \\ $$$${f}\left({x}\right)\:=\:\mathrm{2}\left(\mathrm{sin}\left(\mathrm{1}\right)−\mathrm{1}\right)\mathrm{cos}\left({x}+\mathrm{2}\right) \\ $$$$ \\ $$$$\mathrm{Min}\left({f}\right)\:=\:\mathrm{2}\left(\mathrm{sin}\left(\mathrm{1}\right)−\mathrm{1}\right) \\ $$$$\mathrm{when}\:\mathrm{cos}\left({x}+\mathrm{2}\right)\:=\:\mathrm{1} \\ $$$${x}+\mathrm{2}\:=\:\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi,\:{k}\in\mathbb{Z} \\ $$$${x}\:=\:\frac{\pi}{\mathrm{2}}+\mathrm{2}\left({k}\pi−\mathrm{1}\right),\:{k}\in\mathbb{Z} \\ $$

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