Question

Let $0<\alpha <\frac{\pi }{2}$ be a fixed angle. If $p=(\cos \theta ,\sin \theta )$ and $Q=(\cos (\alpha -\theta ),\sin (\alpha -\theta ))$ then Q is obtained from P by

• A. clockwise rotation around origin through an angle $\tan \alpha$
• B. anticlockwise rotation around origin through an angle $\tan \alpha$
• C. reflection in the line through origin with slop $\tan \alpha$
• D. reflection in the line through origin with slope $\tan \frac{\alpha }{2}$

Answer 1

The correct option is D reflection in the line through origin with slope $\tan \frac{\alpha }{2}$

Clockwise rotation of $P$ through an angle $\alpha$ take it to the point.

$(\cos (\theta -\alpha ),\sin (\theta -\alpha ))$ and anti clockwise take it to $(\cos (\alpha +\theta ),\sin (\alpha +\theta ))$

Now slope of $PQ$
$=\frac{\sin \theta -\sin (\alpha -\theta )}{\cos \theta -\cos (\alpha -\theta )}$

$=\frac{2\cos \left(\frac{\alpha }{2}\right)\sin (\theta -\frac{\alpha }{2})}{-2\sin \left(\frac{\alpha }{2}\right)(-\sin (\theta -\frac{\alpha }{2}))}$

$=-\cot \left(\frac{\alpha }{2}\right)$

$\Rightarrow$ $PQ$ is perpendicular to the line with slope $\tan \left(\frac{\alpha }{2}\right)$

Hence $Q$ is the reflection of $P$ in the line through the origin with

$\tan \left(\frac{\alpha }{2}\right)$