How to show this inequality? 
Show that $$-2 \le \cos \theta(\sin \theta+\sqrt{\sin^2 \theta +3})\le2$$

Trial: I know that $-\dfrac 1 2 \le \cos \theta\cdot\sin \theta \le \dfrac 1 2$ and $\sqrt 3\le\sqrt{\sin^2 \theta +3}\le2$. The problem looks simple to me but I am stuck to solve this. Please help. Thanks in advance.

Essentially, we want the extremum of
$$f(a) = a\sqrt{1-a^2} + a\sqrt{4-a^2}$$
We have
$$f'(a) = \sqrt{1-a^2} + \sqrt{4-a^2} - \dfrac{a^2}{\sqrt{1-a^2}} - \dfrac{a^2}{\sqrt{4-a^2}} = \dfrac{1-2a^2}{\sqrt{1-a^2}} + \dfrac{4-2a^2}{\sqrt{4-a^2}}$$
Let $a^2 = x$, we then want to solve for
\begin{align}\dfrac{1-2x}{\sqrt{1-x}} + \dfrac{4-2x}{\sqrt{4-x}} = 0 \implies (1-2x)^2(4-x) = 4(2-x)^2(1-x)\\
(-4x^3+20x^2-17x+4)-(-4x^3+20x^2-32x+16) = 0 \implies15x-12=0 \implies x = \dfrac45
\end{align}
Hence, $a^*= \pm \dfrac2{\sqrt5}$, which gives us $f(a^*) = \pm 2$.
