## Trigonometry 7th Edition

$y=\sin{(2\cdot0)} = \sin{0}=0$, thus, the point is $(0, 0)$ $y=\sin{(2 \cdot \frac{\pi}{4})} = 1$, thus, the point is $(\frac{\pi}{4}, 1)$ $y=\sin{(2 \cdot \frac{\pi}{2})} =0$, thus, the point is $(\frac{\pi}{2}, 0)$ $y=\sin{(2 \cdot \frac{3\pi}{4})} = -1$, the point is $(\frac{3\pi}{4}, -1)$ $y=\sin{(2\cdot \pi)}=0$, 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 $x=0$: $y=\sin{(2\cdot0)} = \sin{0}=0$, thus, the point is $(0, 0)$ When $x=\frac{\pi}{4}$: $y=\sin{(2 \cdot \frac{\pi}{4})} = \sin{\frac{\pi}{2}} =1$, thus, the point is $(\frac{\pi}{4}, 1)$ When $x=\frac{\pi}{2}$: $y=\sin{(2 \cdot \frac{\pi}{2})} =\sin{\pi} = 0$, thus, the point is $(\frac{\pi}{2}, 0)$ When $x=\frac{3\pi}{4}$: $y=\sin{(2\cdot \frac{3\pi}{4})} = \sin{\frac{3\pi}{2}}$ The reference angle is $\frac{\pi}{2}$. Since $\frac{3\pi}{2}$ is on the negative y-axis, sine is negative. Thus, $y=\sin{\frac{3\pi}{2}} = -\sin{\frac{\pi}{2}} = -1$ The point is $(\frac{3\pi}{4}, -1)$ When $x=\pi$: $y=\sin{(2\cdot \pi)}=\sin{2\pi} = 0$ Thus, the point is $(\pi, 0)$