$\text{ Let }0<\alpha <\pi /2\text{ be a fixed angle. If }P\equiv (\cos \theta ,\sin \theta )\text{ and }Q\equiv (\cos (\alpha -\theta ),\sin (\alpha -\theta \left)\right),\text{ then }Q\text{ is obtained from }P by the$

$\text{ Clearly, points }P(\cos \theta ,\sin \theta )\text{ and }Q(\cos (\alpha -\theta )\sin (\alpha -\theta )\text{ ) lie on circle of unit radius. }$

$\text{ In the figure, }\mathrm{\angle }POX=\theta \text{ and }\mathrm{\angle }QOX=\alpha -\theta \text{ . }$

$\begin{array}{r}\therefore \mathrm{\angle }QOP=\alpha -2\theta \\ \text{ Now, }\mathrm{\Delta }QOP\text{ is isosceles. }\end{array}$

Therefore, altitude or angle bisector OM is perpendicular bisector of PQ. 

$\begin{array}{r}\mathrm{\angle }MOP=\frac{\alpha -2\theta }{2}=\frac{\alpha }{2}-\theta \\ \mathrm{\angle }MOX=\theta +\left(\frac{\alpha }{2}-\theta \right)=\frac{\alpha }{2}\end{array}$

Thus, point P is reflection of point Q inline OM having slope, tan $\left(\frac{\alpha }{2}\right)$