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Find-minimum-value-of-f-x-sin-x-3-sin-x-1-2cos-x-2-where-x-R-

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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
Findminimumvalueoff(x)=sin(x+3)−sin(x+1)−2cos(x+2)wherexϵR
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)) :}
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)−2max=−2sin(1)+2
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(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)whencos(x+2)=1x+2=π2+2kπ,k∈Zx=π2+2(kπ−1),k∈Z

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