## Trigonometry 7th Edition

$y=-\sin{0} = 0$, thus, the point is $(0, 0)$ $y=-\sin{\frac{\pi}{2}} = -1$, thus, the point is $(\frac{\pi}{2}, -1)$ $y=-\sin{\pi} = 0$, thus, the point is $(\pi, 0)$ $y=-\sin{\frac{3\pi}{2}} = \sin{\frac{\pi}{2}}=1$, thus, the point is $(\frac{3\pi}{2}, 1)$. $y=-\sin{2\pi} =0$, thus, the point is $(2\pi, 0)$
All the given values of $x$ are either special angles or have a special angle as their reference angle. Hence, the value of the cosine 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 $x=\frac{\pi}{2}$: $y=-\sin{\frac{\pi}{2}} = -1$, thus, the point is $(\frac{\pi}{2}, -1)$ When $x=\pi$: $y=-\sin{\pi} = 0$, thus, the point is $(\pi, 0)$ When $x=\frac{3\pi}{2}$: Reference angle is $\frac{\pi}{2}$. Since the angle is on the negative y-axis, then sine is negative. Thus, $y=-\sin{\frac{3\pi}{2}} =-( -\sin{\frac{\pi}{2}})=-(-1)=1$ Thus, the point is $(\frac{3\pi}{2}, 1)$. When $x=2\pi$: $y=-\sin{2\pi} =0$, thus, the point is $(2\pi, 0)$