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

Jul 25, 2018

$f \left(x\right) = \left(3 \sin x - 4 \cos x - 10\right) \left(3 \sin x + 4 \cos x - 10\right)$

$= \left(\left(3 \sin x - 10\right) - 4 \cos x\right) \left(\left(3 \sin x - 10\right) + 4 \cos x\right)$

$= {\left(3 \sin x - 10\right)}^{2} - {\left(4 \cos x\right)}^{2}$

$= 9 {\sin}^{2} x - 60 \sin x + 100 - 16 {\cos}^{2} x$

$= 9 {\sin}^{2} x - 60 \sin x + 100 - 16 + 16 {\sin}^{2} x$

$= 25 {\sin}^{2} x - 60 \sin x + 84$

$= {\left(5 \sin x\right)}^{2} - 2 \cdot 5 \sin x \cdot 6 + {6}^{2} - {6}^{2} + 84$

$= {\left(5 \sin x - 6\right)}^{2} + 48$

$f \left(x\right)$ will be maximum when ${\left(5 \sin x - 6\right)}^{2}$ is maximum . It will be possible for $\sin x = - 1$

So

${\left[f \left(x\right)\right]}_{\text{max}} = {\left(5 \left(- 1\right) - 6\right)}^{2} + 48 = 169$

Jul 25, 2018

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

#### Explanation:

Let $\alpha = {\sin}^{- 1} \left(\frac{4}{5}\right)$..Then

$f \left(x\right) = 25 \left(\sin \left(x - \alpha\right) - 2\right) \left(\sin \left(x + \alpha\right) - 2\right)$
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]}