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

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

Find minimum value of

f (x) = \sin (x + 3) - \sin (x + 1) - 2 \cos (x + 2)

where x ϵ R

Answered by EDWIN88 last updated on 25/Jun/21

f (x) = \sin (x + 3) - \sin (x + 1) - 2 \cos (x + 2)

f (x) = 2 \cos (x + 2) \sin (1) - 2 \cos (x + 2)

f (x) = 2 (\sin (1) - 1) \cos (x + 2)

f (x) = {\begin{cases} \min = 2 \sin (1) - 2 \\ \max = - 2 \sin (1) + 2 \end{cases}

Answered by Olaf_Thorendsen last updated on 25/Jun/21

f (x) = \sin (x + 3) - \sin (x + 1) - 2 \cos (x + 2)

f (x) = 2 \sin (\frac{(x + 3) - (x + 1)}{2}) \cos (\frac{(x + 3) + (x + 1)}{2})

- 2 \cos (x + 2)

f (x) = 2 \sin (1) \cos (x + 2) - 2 \cos (x + 2)

f (x) = 2 (\sin (1) - 1) \cos (x + 2)

Min (f) = 2 (\sin (1) - 1)

when \cos (x + 2) = 1

x + 2 = \frac{π}{2} + 2 k π, k \in Z

x = \frac{π}{2} + 2 (k π - 1), k \in Z