Question

Find the maximum and minimum values of $f\left(x\right)=\cos (x+\frac{\pi }{3})+2\sqrt{2}\sin (x+\frac{\pi }{3})-3$

Answer 1

$f\left(x\right)=\cos (x+\frac{\pi }{3})+2\sqrt{2}\sin (x+\frac{\pi }{3})-3$

For $a\cos x+b\sin x$ we know that the maximum and minimum values are $+\sqrt{a^2+b^2}$ and $-\sqrt{a^2+b^2}$

Thus, for $f\left(x\right)$, the maximum value is

$\sqrt{{\left(2\sqrt{2}\right)}^2+1^2}-3$

$=\sqrt{8+1}-3$

$=3-3$

$=0$

The minimum value for $f\left(x\right)$

$=-\sqrt{{\left(2\sqrt{2}\right)}^2+1^2}-3$

$=-\sqrt{8+1}-3$

$=-3-3$

$=-6$

The maximum value of $f\left(x\right)$ is $0$ and the minimum value is $-6$