trigonometry - How can we bound $frac{sin(theta)cos(theta)}{|cos(theta)|+|sin(theta)|}$

How can we bound $\frac{\sin(\theta)\cos(\theta)}{|\cos(\theta)|+|\sin(\theta)|}$ from above and below? I can see that it is between $\frac{-1}{2}$ and $\frac{1}{2}$ by plotting it. I'm using this to show that the function
\begin{align}
f(x,y)=
\begin{cases}
\frac{xy}{|x|+|y|}\text{if $(x,y)\neq (0,0)$}\\
0\text{ else}
\end{cases}
\end{align}

is continuous by converting to polar coordinates.

Answer

We know
$$(|cos \theta| + |sin \theta|)^2 = 1 + 2|cos \theta sin \theta| = 1 + |sin (2\theta)|$$

So
$$ \frac{\sin(\theta)\cos(\theta)}{|\cos(\theta)|+|\sin(\theta)|} = \frac{sin(2\theta)}{2\sqrt{1+|sin(2\theta)|}}$$

So the bound is given by the maximum of $f(x) = \frac{x}{2\sqrt{1 +x}}$ for $0\le x \le 1$. F(x) is monotonne increasing, so the maximum is $f(1) = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$

Therefore,
$$|\frac{\sin(\theta)\cos(\theta)}{|\cos(\theta)|+|\sin(\theta)|}| \le \frac{\sqrt{2}}{4}$$

The equality holds at $\theta = \frac{\pi}{4}$