## Trigonometry 7th Edition

$y=-\cos{0} = -1$, thus, the point is $(0, -1)$ $y=-\cos{\frac{\pi}{2}} = 0$, thus, the point is $(\frac{\pi}{2}, \frac{\sqrt2}{2})$ $y=-\cos{\pi} = 1$, thus, the point is $(\pi, 1)$ $y=-\cos{\frac{3\pi}{2}} = -\cos{\frac{\pi}{2}}=0$, thus, the point is $(\frac{3\pi}{2}, 0)$ $y=-\cos{2\pi} =-1$, thus, the point is $(2\pi, -1)$
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 $x=0$: $y=-\cos{0} = -1$, thus, the point is $(0, -1)$ When $x=\frac{\pi}{2}$: $y=-\cos{\frac{\pi}{2}} = 0$, thus, the point is $(\frac{\pi}{2}, 0)$ When $x=\pi$: $y=-\cos{\pi} = -(-1)=1$, thus, the point is $(\pi, 1)$ When $x=\frac{3\pi}{2}$: Reference angle is $\frac{\pi}{2}$. Since the angle is on the negative y-axis cosine is zero. Thus, $y=-\cos{\frac{3\pi}{2}} = -\cos{\frac{\pi}{2}}=0$ Thus, the point is $(\frac{3\pi}{2}, 0)$. When $x=2\pi$: $y=-\cos{2\pi} =-1$, thus, the point is $(2\pi, -1)$