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 $\displaystyle -\sqrt{1+\sin^2(\alpha)}\leq f(x)\leq \sqrt{1+\sin^2(\alpha)}$

Let $p=\sin^2\alpha,\; y=\cos 2x,\;$ then
$$\cos x \sin x+\cos x\sqrt{\sin^2x+\sin^2\alpha\mathstrut}
=\dfrac{\pm\sqrt{1-y^2}\pm\sqrt{1-y^2+2p(1+y)\mathstrut}}2 = f(y),$$
$$2f'(y) = \mp\dfrac y{\sqrt{1-y^2}}\pm\dfrac{p-y}{\sqrt{1-y^2+2p(1+y)}}.$$
The set of possible extremes of $f(y)$ corresponds with the expression
$$\left[\begin{align}
&y^2-1=0\\[4pt]
&y^2-2py-2p-1=0\\[4pt]
&y^2(1+2py-y^2+2p)=(p^2-2py+y^2)(1-y^2),
\end{align}\right.,$$
$$\left[\begin{align}
&y=\pm1\\[4pt]
&y=p\pm\sqrt{p^2+2p+1}\\[4pt]
&(2+p)y^2+2y-p=0,
\end{align}\right.$$
$$y_m\in\left\{-1,1, \dfrac p{2+p}\right\},\quad
f(y_m)\in\left\{0, \pm\sqrt{p\mathstrut}, \pm \sqrt{1+p} \right\},$$
$$\mathbb{f(y)\in\left[-\sqrt{1+\sin^2\alpha}, \sqrt{1+\sin^2\alpha}\right]}.$$
