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