Answer
$y=-\cos{0} = -1$, 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 $(2\pi, -1)$
Work Step by Step
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 $(2\pi, -1)$
