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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.

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  • $\begingroup$ 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 $Y$? $\endgroup$
    – Taladris
    Mar 20 at 4:07

1 Answer 1

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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 $\pi^2$.

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