Question

$$a<c\cos(-\frac{1}{2})+a\ rc\sin(-\frac{\sqrt{2}}{2})=$$

Solve for r

$\left\{\begin{matrix}r<\frac{\cos(\frac{1}{2})c-a}{\sin(\frac{\sqrt{2}}{2})ac}\text{, }&\left(c<0\text{ and }a<0\right)\text{ or }\left(c>0\text{ and }a>0\right)\\r\in \mathrm{R}\text{, }&\left(a<0\text{ and }c=0\right)\text{ or }\left(c>0\text{ and }a=0\right)\\r>\frac{\cos(\frac{1}{2})c-a}{\sin(\frac{\sqrt{2}}{2})ac}\text{, }&\left(c>0\text{ and }a<0\right)\text{ or }\left(a>0\text{ and }c<0\right)\end{matrix}\right.$

Solve for c

$\left\{\begin{matrix}c<\frac{a}{-\sin(\frac{\sqrt{2}}{2})ar+\cos(\frac{1}{2})}\text{, }&\left(a>\frac{\cos(\frac{1}{2})}{\sin(\frac{\sqrt{2}}{2})r}\text{ and }r>0\right)\text{ or }\left(a<\frac{\cos(\frac{1}{2})}{\sin(\frac{\sqrt{2}}{2})r}\text{ and }r<0\right)\\c\in \mathrm{R}\text{, }&a=\frac{\cos(\frac{1}{2})}{\sin(\frac{\sqrt{2}}{2})r}\text{ and }r<0\\c>\frac{a}{-\sin(\frac{\sqrt{2}}{2})ar+\cos(\frac{1}{2})}\text{, }&\left(r>0\text{ and }a<\frac{\cos(\frac{1}{2})}{\sin(\frac{\sqrt{2}}{2})r}\right)\text{ or }r=0\text{ or }\left(a>\frac{\cos(\frac{1}{2})}{\sin(\frac{\sqrt{2}}{2})r}\text{ and }r<0\right)\end{matrix}\right.$