Question

The maximum value of `f(x)=(3sinx-4cosx-10)(3sinx+4cosx-10)` is?

Answer 1

`f(x)=(3sinx-4cosx-10)(3sinx+4cosx-10)`

`=((3sinx-10)-4cosx)((3sinx-10)+4cosx)`

`=(3sinx-10)^2-(4cosx)^2`

`=9sin^2x-60sinx+100-16cos^2x`

`=9sin^2x-60sinx+100-16+16sin^2x`

`=25sin^2x-60sinx+84`

`=(5sinx)^2-2*5sinx*6+6^2-6^2+84`

`=(5sinx-6)^2+48`

`f(x)` will be maximum when `(5sinx-6)^2` is maximum . It will be possible for `sinx=-1`

So

`[f(x)]_"max"=(5(-1)-6)^2+48=169`

Answer 2

Maximum is 169. Minimum is 50 (perhaps, nearly). This is graphical illustration, for Dilip's answer.

Explanation:

Let `alpha = sin^(-1) ( 4/5 )`..Then

`f( x ) = 25 (sin ( x - alpha )-2)(sin (x + alpha) - 2)`
See graph.
graph{(y - 25 (sin (x- 0.9273)-2)(sin (x+ 0.9273)-2))(y-169)(y-50)=0[-20 20 20 230]}
graph{(y - 25 (sin (x- 0.9273)-2)(sin (x+ 0.9273)-2))(y-169)=0[-1.75 -1.5 167 171]}