Let 0

Question

(D) The point $P$ makes an angle $	heta$ with the positive $x$-axis. The point $Q$ makes an angle $(\alpha - 	heta)$ with the positive $x$-axis. The

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By reflection in a line passing through the origin with slope $	an(\alpha/2)$.

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(D) The point $P$ makes an angle $	heta$ with the positive $x$-axis. The point $Q$ makes an angle $(\alpha - 	heta)$ with the positive $x$-axis. The angle between $OP$ and $OQ$ is $\angle POQ = 	heta - (\alpha - 	heta) = 2	heta - \alpha$ is incorrect; rather,the angle between the vectors is $	heta - (\alpha - 	heta)$ is not the approach. Let us consider the rotation: $A$ rotation of $P(\cos 	heta, \sin 	heta)$ by an angle $\phi$ clockwise results in $(\cos(	heta - \phi), \sin(	heta - \phi))$. Setting $	heta - \phi = \alpha - 	heta$,we get $\phi = 2	heta - \alpha$,which depends on $	heta$. Re-evaluating: The reflection of $P(\cos 	heta, \sin 	heta)$ in a line $y = x 	an(\alpha/2)$ (which makes an angle $\alpha/2$ with the $x$-axis) is given by the rotation formula for reflection: $x' = \cos(2\phi - 	heta), y' = \sin(2\phi - 	heta)$ where $\phi = \alpha/2$. Thus,$x' = \cos(\alpha - 	heta)$ and $y' = \sin(\alpha - 	heta)$. Therefore,$Q$ is the reflection of $P$ in the line passing through the origin with slope $	an(\alpha/2)$.

Let $0 < \alpha < \pi/2$ be a constant angle. If $P \equiv (\cos \theta, \sin \theta)$ and $Q \equiv (\cos(\alpha - \theta), \sin(\alpha - \theta))$,how is $Q$ obtained from $P$?

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