$\forall x\in R,$find the range of the function $f(x)=\cos x(\sin x+\sqrt{\sin^2x+\sin^2\alpha});\alpha\in[0,\pi]$
$f(x)=\cos x(\sin x+\sqrt{\sin^2x+\sin^2\alpha});\alpha\in[0,\pi]$
$f'(x)=\cos x(\cos x+\frac{\sin x\cos x}{\sqrt{\sin^2x+\sin^2\alpha}})-\sin x(\sin x+\sqrt{\sin^2x+\sin^2\alpha})$
I am stuck here and could not find the minimum and maximum values of $f(x),$The answer given is $-\sqrt{1+\sin^2\alpha}\leq f(x)\leq\sqrt{1+\sin^2\alpha}$.
Best Answer
Let $$y=\cos x\left[\sin x+\sqrt{\sin^2 x+\sin^2 a}\right] = \sin x\cdot \cos x+\cos x\cdot \sqrt{\sin^2 x+\sin^2 a}$$
Now Using $\bf{Cauchy\; Schwartz\; Inequality}$
We get $$(\sin^2 x+\cos ^2 x)\cdot \left[\cos^2 x+\sin^2 x+\sin^2 a\right]\geq \left(\sin x\cdot \cos x+\cos x\cdot \sqrt{\sin^2 x+\sin^2 a}\right)^2$$
So we get $$y^2\leq (1+\sin^2 a)\Rightarrow |y| \leq\sqrt{1+\sin^2 a}$$
So we get $$-\sqrt{1+\sin^2 a}\leq y \leq \sqrt{1+\sin^2 a}.$$