# If $x,y∈(-π,π]$, then find the area of the polygon formed by points $(x,y)$ satisfying the equation $\lfloor|\sin x|\rfloor+\lfloor|\cos y|\rfloor=2$.

If $$x,y∈(-π,π]$$, then find the area of the polygon formed by points $$(x,y)$$ satisfying the equation $$\lfloor|\sin x|\rfloor+\lfloor|\cos y|\rfloor=2$$.

My attempts include using a graphing tool and taking $$\sin x= \pm1$$ and $$\cos y=\pm1$$ for $$x=\pm \displaystyle{\frac{\pi}{2}}$$ and $$y=0,\pi$$. How do you solve this further? $$\lfloor\,\cdot\,\rfloor$$ represents the greatest integer function / floor function.

• Let $X=|\sin(x)|$ and $Y=|\cos(y)|$. Using that $-1\le\sin(x)\le 1$ and $-1\le\cos(y)\le 1$, what are the possible values for $X$ and $Y$? Mar 20 at 4:07

You are on the right track. You know that the equation is satisfied for $$x \in \{-\frac{\pi}{2},\frac{\pi}{2}\}$$ and $$y \in \{0, \pi\}$$. So if you take the cartesian product of those two sets you have the list of all the points satisfying this equation : $$\{(-\frac{\pi}{2}, 0),\; (-\frac{\pi}{2}, \pi),\;(\frac{\pi}{2}, 0),\; (\frac{\pi}{2}, \pi)\}$$. You can see by graphing, or simply by looking at the coordinates that this forms a square of side $$\pi$$. So the area you are looking for is $$\pi^2$$.