Question

The triangles OAB is right angles when points O , A , B are (0 , 0) , ( $cos\theta ,sin\theta )$ and $(cos\varphi ,sin\varphi )$ respectively , then $\theta and\varphi$ are connected by the relations
$sin\left(\frac{\theta -\varphi }{2}\right)=\pm \frac{1}{\sqrt{2}}$and $cos\left(\frac{\theta -\varphi }{2}\right)=\pm \frac{1}{\sqrt{2}}$

Answer 1

Clearly OA = OB = 1 as $co{s}^{2}x+si{n}^{2}x=1$
$\therefore A{B}^2=O{A}^2+O{B}^2=2$
or ${(cos\theta -cos\varphi )}^2+{(sin\theta -sin\varphi )}^2=2$
or $1+1+2(cos\theta \cdot cos\varphi +sin\theta \cdot sin\varphi )=2$
or $cos(\theta -\varphi )=0or2co{s}^2\frac{\theta -\varphi }{2}-1=0$
or $1-2si{n}^2\frac{\theta -\varphi }{2}=0$
All the four relations follow from above