I'm working on a larger proof, but need to show that the following has no solutions. $$0 \le x,y < 2\pi$$ $$ R\cos(x)-C\cos(x) = R\cos(y)+C\cos(y)\\ R\sin(x)-C\sin(x) = R\sin(y)+C\sin(y) $$ I've tried for a few hours, but have not found a concise method to show this is true.
I know already that the following can yield a solution, but I'd like to avoid that if possible. $$ C(-\cos(\frac{x}{2} + \frac{x}{2}) - \cos(\frac{y}{2} + \frac{y}{2}) = R(\cos(\frac{y}{2} + \frac{y}{2}) - \cos(\frac{x}{2} + \frac{x}{2}))\\ C(-\sin(\frac{x}{2} + \frac{x}{2}) - \sin(\frac{y}{2} + \frac{y}{2}) = R(\sin(\frac{y}{2} + \frac{y}{2}) - \sin(\frac{x}{2} + \frac{x}{2})) $$ Simplifes to $$ -2C\cos(\frac{x}{2} - \frac{y}{2}) \cos(\frac{x}{2} + \frac{y}{2}) = 2R\sin(\frac{x}{2} - \frac{y}{2}) \sin(\frac{x}{2} + \frac{y}{2})\\ 2C\sin(\frac{x}{2} + \frac{y}{2}) \cos(\frac{x}{2} - \frac{y}{2}) = 2R\sin(\frac{x}{2} - \frac{y}{2}) \cos(\frac{x}{2} + \frac{y}{2}) $$