The difference of the max and the min values of the function $5\cos x+3\cos (x+\frac{\pi }{3})+8$ over R is

Let $f\left(x\right)=5\cos x+3\cos (x+60°)+8$

$=5\cos x+3(\cos x\cos 60°-\sin x\sin 60°)+8$

$=\frac{13}{2}\cos x-\frac{3\sqrt{3}}{2}\sin x+8$

This is of the form $a\cos x+b\sin x+c,$

where $a=\frac{13}{2},b=\frac{-3\sqrt{3}}{2},c=8$

We have $\sqrt{a^2+b^2}=7$

The extreme values of f over R are

$c\pm \sqrt{a^2+b^2}=8\pm 7=15,1$

‘15’ is the max value and ‘1' is the min value of the function f over R