## Precalculus (6th Edition) Blitzer

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.