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 $\displaystyle -\sqrt{1+\sin^2(\alpha)}\leq f(x)\leq \sqrt{1+\sin^2(\alpha)}$
 A: I'm going to have a look on your try, and then add a claim and an example which might help you.
From
$$-(\cos^2x+\sin^2x)\leqslant 2\cos x\sin x\leqslant \cos^2x+\sin^2x\tag1$$
and
$$\small -(\cos^2 x+\sin^2 x+\sin^2\alpha)\leqslant 2\cos x\sqrt{\sin^2x+\sin^2\alpha}\leqslant \cos^2x+\sin^2x+\sin^2\alpha\tag2$$
you got
$$-2-\sin^2\alpha\leqslant 2f(x)\leqslant 2+\sin^2\alpha$$
and
$$-1-\frac{\sin^2\alpha}{2}\leqslant f(x)\leqslant 1+\frac{\sin^2\alpha}{2}\tag3$$
To see if $(3)$ is the range of $f(x)$, let us see if there is $x$ such that, for any given $\alpha$, $f(x)=1+\dfrac{\sin^2\alpha}{2}$, i.e.
$$2f(x)=2+\sin^2\alpha\tag4$$
holds.
From $(1)(2)$, $x$ satisfying $(4)$ has to satisfy both
$$2\cos x\sin x= \cos^2x+\sin^2x\tag5$$
and
$$2\cos x\sqrt{\sin^2x+\sin^2\alpha}=\cos^2x+\sin^2x+\sin^2\alpha\tag6$$
$(5)$ is equivalent to
$$(\cos x-\sin x)^2=0\iff \cos x=\sin x\iff x=\frac{\pi}{4}+m\pi\ (m\in\mathbb Z)$$
$(6)$ is equivalent to
$$\cos x\geqslant 0\quad\text{and}\quad \cos^2x=\sin^2x+\sin^2\alpha\iff \cos x\geqslant 0\quad\text{and}\quad \cos(2x)=\sin^2\alpha$$
So, we see that
$$(5)\ \ \text{and}\ \ (6)\iff x=\dfrac{\pi}{4}+2m\pi\quad\text{and}\quad \alpha=n\pi\quad (m,n\in\mathbb Z)$$
This means that there is no $x$ such that, for any given $\alpha$, $f(x)=1+\dfrac{\sin^2\alpha}{2}$ holds.
So, $(3)$ is not the range of $f(x)$.

In the following, I'm going to add a claim and an example which might help you.
In the following, I consider only continuous functions.
Claim : If

*

*$f(x)\leqslant F(x)\leqslant g(x)\quad$ ($f(x),g(x)$ are not necessarily constant functions)


*$h(x)\leqslant G(x)\leqslant i(x)\quad$ ($h(x),i(x)$ are not necessarily constant functions)


*both $f(x)+h(x)$ and $g(x)+i(x)$ are constant functions


*there is $\alpha$ such that $F(\alpha)=f(\alpha)$ and $G(\alpha)=h(\alpha)$


*there is $\beta$ such that $F(\beta)=g(\beta)$ and $G(\beta)=i(\beta)$
then, $\operatorname{range}(F(x)+G(x))=\left[f(x)+h(x),g(x)+i(x)\right]$.
Proof :
For $c$ such that $f(x)+h(x)\leqslant c\leqslant g(x)+i(x)$, letting $H(x)=F(x)+G(x)-c$, since we have
$$H(\alpha)=F(\alpha)+G(\alpha)-c=f(\alpha)+h(\alpha)-c\leqslant 0$$
$$H(\beta)=F(\beta)+G(\beta)-c=g(\beta)+i(\beta)-c\geqslant 0$$
we can say that, by the intermediate value theorem, there is $\gamma$ such that
$$H(\gamma)=0\qquad \text{and}\qquad \min(\alpha,\beta)\leqslant \gamma\leqslant \max(\alpha,\beta).\quad\blacksquare$$

Exmaple for which the claim above works :
Find the range of $\cos x+\dfrac{x^2+8\pi x-2\pi^2}{3x^2+6\pi^2}$.
Let $f(x)=\cos x$ and $g(x)=\dfrac{x^2+8\pi x-2\pi^2}{3x^2+6\pi^2}$. Then, $g'(x)=\dfrac{-8 \pi (x-2\pi)(x+\pi)}{3 (x^2 + 2 \pi^2)^2}$ and $\displaystyle\lim_{x\to\pm\infty}g(x)=\frac 13$.
Since $-1\leqslant f(x)\leqslant 1$ and $-1=g(-\pi)\leqslant g(x)\leqslant g(2\pi)=1$, adding these gives
$$-2\leqslant f(x)+g(x)\leqslant 2\tag7$$
Since $f(-\pi)=g(-\pi)=-1$ and $f(2\pi)=g(2\pi)=1$, we can say, by the claim above, that $(7)$ is the range.
A: 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
$$\operatorname{range}(f)=\left[-\sqrt{1+a^2},\sqrt{1+a^2}\right].$$
A: 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
$$\left[\begin{align}
&y^2-1=0\\[4pt]
&y^2-2py-2p-1=0\\[4pt]
&y^2(1+2py-y^2+2p)=(p^2-2py+y^2)(1-y^2),
\end{align}\right.,$$
$$\left[\begin{align}
&y=\pm1\\[4pt]
&y=p\pm\sqrt{p^2+2p+1}\\[4pt]
&(2+p)y^2+2y-p=0,
\end{align}\right.$$
$$y_m\in\left\{-1,1, \dfrac p{2+p}\right\},\quad
f(y_m)\in\left\{0, \pm\sqrt{p\mathstrut}, \pm \sqrt{1+p} \right\},$$
$$\mathbb{f(y)\in\left[-\sqrt{1+\sin^2\alpha}, \sqrt{1+\sin^2\alpha}\right]}.$$
