Question

The minimum value of function $\left(\sin \right({\sin }^{-1}x){)}^2-6\sin ({\sin }^{-1}x)+10\left(3\right({\sin }^{4}x+{\cos }^{4}x)-2({\sin }^{6}x+{\cos }^{6}x\left)\right),x\in [-1,1]$ is

Answer 1

Let $f\left(x\right)=\left(\sin \right({\sin }^{-1}x){)}^2-6\sin ({\sin }^{-1}x)+10(3({\sin }^{4}x+{\cos }^{4}x)-2({\sin }^{6}x+{\cos }^{6}x))$
Domain of $f\left(x\right)$ is $[-1,1]$
Now,
$3({\sin }^{4}x+{\cos }^{4}x)-2({\sin }^{6}x+{\cos }^{6}x)$
$\Rightarrow 3({\sin }^{4}x+{\cos }^{4}x)-2\left(\right({\sin }^{2}x+{\cos }^{2}x\left)\right({\sin }^4+{\cos }^4-{\sin }^{2}x{\cos }^{2}x\left)\right)$
$\Rightarrow {\sin }^4+{\cos }^4+2{\sin }^{2}x{\cos }^{2}x$
$\Rightarrow ({\sin }^{2}x+{\cos }^{2}x{)}^2=1$
and $\sin \left({\sin }^{-1}x\right)=x\forall x\in [-1,1]$
$\therefore f\left(x\right)=x^2-6x+10=(x-3{)}^2+1$
$\Rightarrow 5\le f\left(x\right)\le 17(\because x\in [-1,1\left]\right)$
$\therefore$ Minimum value of $f\left(x\right)$ is $5.$