# 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.

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 $\theta = \frac{\pi}{4}$