## Trigonometry 7th Edition

$y=\frac{1}{2}\cos{0} =\frac{1}{2}$, 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 $(2\pi, \frac{1}{2})$
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 $(2\pi, \frac{1}{2})$