## Precalculus (6th Edition) Blitzer

The statement is false because the trigonometric function is defined in all quadrant angles. For a quadrant angle $\theta =0$ , $\sin 0=0$ For a quadrant angle $\theta =\frac{\pi }{2}$ , $\sin \frac{\pi }{2}=1$ For a quadrant angle $\theta =\pi$ , $\sin \pi =0$ For a quadrant angle $\theta =\frac{3\pi }{2}$ , $\sin \frac{3\pi }{2}=-1$ Thus, the statement does not make sense.