Answer
The statement does not make sense..
Work Step by Step
This, statement does not make sense. Consider an example.
Let us suppose we want to find the value of $\sin \left( {{112.5}^{\circ }} \right)$. Because $112.5{}^\circ =\frac{225}{2}$, the half angle formula for $\sin \frac{\alpha }{2}$ with $\alpha =225{}^\circ $ can be used:
$\sin \frac{\alpha }{2}=\pm \sqrt{\frac{1-\cos \alpha }{2}}$
The value can be computed as given below:
$\begin{align}
& \cos 225{}^\circ =\cos \left( 180{}^\circ +45{}^\circ \right) \\
& =-\cos \left( 45{}^\circ \right) \\
& =-\frac{1}{\sqrt{2}}
\end{align}$
Therefore, by using the aforementioned formula:
$\begin{align}
& \sin \left( {{112.5}^{\circ }} \right)=\pm \sqrt{\frac{1-\left( -\frac{1}{\sqrt{2}} \right)}{2}} \\
& =\pm \frac{\sqrt{2-\sqrt{2}}}{2}
\end{align}$
$112.5{}^\circ $ lies in the second quadrant, therefore, $\sin \left( 112.5{}^\circ \right)$ will be positive.
$\sin \left( {{112.5}^{\circ }} \right)=\frac{\sqrt{2-\sqrt{2}}}{2}$
Thus, the sign of the half angle is determined by the quadrant where the half angle $\frac{\alpha }{2}$ lies. So, the angle and its half do not necessarily lie on the same quadrant.
