<s>[INST] Range of Trigonometric function having square root 
Finding range of function


$\displaystyle f(x)=\cos(x)\sin(x)+\cos(x)\sqrt{\sin^2(x)+\sin^2(\alpha)}$

I have use Algebric inequality
$\displaystyle -(a^2+b^2)\leq 2ab\leq (a^2+b^2)$
$\displaystyle (\cos^2(x)+\sin^2(x))\leq 2\cos(x)\sin(x)\leq \cos^2(x)+\sin^2(x)\cdots (1)$
And
$\displaystyle -[\cos^2(x)+\sin^2(x)+\sin^2(\alpha)]\leq 2\cos(x)\sqrt{\sin^2(x)+\sin^2(\alpha)}\leq [\cos^2(x)+\sin^2(x)+\sin^2(\alpha)]\cdots (2)$
Adding $(1)$ and $(2)$
$\displaystyle -(1+1+\sin^2(\alpha))\leq 2\cos(x)[\sin(x)+\sqrt{\sin^2(x)+\sin^2(\alpha)}]\leq (1+1+\sin^2(\alpha))$
$\displaystyle -\bigg(1+\frac{\sin^2(\alpha)}{2}\bigg)\leq f(x)\leq \bigg(1+\frac{\sin^2(\alpha)}{2}\bigg)$
I did not know where my try is wrong.
Please have a look on it
But actual answer is $\displaystyle -\sqrt{1+\sin^2(\alpha)}\leq f(x)\leq \sqrt{1+\sin^2(\alpha)}$
 [/INST] Let $a=|\sin\alpha|.$
First notice that (since $|\sin x|\le\sqrt{a^2+\sin^2x}$) when the maximum value $f_{\max}$ is attained the $\sin$ and $\cos$ of $x$ are positive, and that the minimum value is $f_{\min}=-f_{\max},$ corresponding to the same $\sin$ and the opposite $\cos.$
$$f'(x)=\left(1+\frac{\sin x}{\sqrt{a^2+\sin^2x}}\right)g(x)$$
where
$$g(x):=\cos^2x-\sqrt{a^2+\sin^2x}\,\sin x$$
decreases continuously from $g(0)=1-0=1$ to $g(\pi/2)=0-a=-a$ hence vanishes for a unique $x_0\in[0,\pi/2].$ One easily checks that
$$\sin^2x_0=\frac1{2+a^2},\quad\cos^2x_0=\frac{1+a^2}{2+a^2}.$$
Therefore, $f_\max=f(x_0)=\sqrt{1+a^2}$ and
$$\operatorname{range}(f)=\left[-\sqrt{1+a^2},\sqrt{1+a^2}\right].$$
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