## Trigonometry 7th Edition

$y=\sin{0} = 0$, thus, the point is $(0, 0)$ $y=\sin{\frac{\pi}{4}} = \frac{\sqrt2}{2}$, thus, the point is $(\frac{\pi}{4}, \frac{\sqrt2}{2})$ $y=\sin{\frac{\pi}{2}} = 1$, thus, the point is $(\frac{\pi}{2}, 1)$ $y=\sin{\frac{3\pi}{4}} = \frac{\sqrt2}{2}$, thus, the point is $(\frac{3\pi}{4}, \frac{\sqrt2}{2})$ $y=\sin{\pi} = 0$, thus, the point is $(\pi, 0)$
All the given values of $x$ are either special angles or have a special angle as their reference angle. Hence, the value of the sine for each angle can be easily found. Evaluate the function for each given value of $x$ to obtain: When $x=0$: $y=\sin{0} = 0$, thus, the point is $(0, 0)$ When $x=\frac{\pi}{4}$: $y=\sin{\frac{\pi}{4}} = \frac{\sqrt2}{2}$, thus, the point is $(\frac{\pi}{4}, \frac{\sqrt2}{2})$ When $x=\frac{\pi}{2}$: $y=\sin{\frac{\pi}{2}} = 1$, thus, the point is $(\frac{\pi}{2}, 1)$ When $x=\frac{3\pi}{4}$: Reference angle is $\frac{\pi}{4}$. Since the angle is in Quadrant II, sine is positive. Thus, $y=\sin{\frac{3\pi}{4}} = \sin{\frac{\pi}{4}}=\frac{\sqrt2}{2}$ Thus, the point is $(\frac{3\pi}{4}, \frac{\sqrt2}{2})$ When $x=\pi$: $y=\sin{\pi} = 0$, thus, the point is $(\pi, 0)$