Trigonometry 7th Edition

$y=\cos{0} = 1$, thus, the point is $(0, 1)$ $y=\cos{\frac{\pi}{4}} = \frac{\sqrt2}{2}$, thus, the point is $(\frac{\pi}{4}, \frac{\sqrt2}{2})$ $y=\cos{\frac{\pi}{2}} = 0$, thus, the point is $(\frac{\pi}{2}, 0)$ $y=\cos{\frac{3\pi}{4}} -\frac{\sqrt2}{2}$, thus, the point is $(\frac{3\pi}{4}, -\frac{\sqrt2}{2})$ $y=\cos{\pi} = -1$, thus, the point is $(\pi, -1)$
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 $x=\frac{\pi}{4}$: $y=\cos{\frac{\pi}{4}} = \frac{\sqrt2}{2}$, thus, the point is $(\frac{\pi}{4}, \frac{\sqrt2}{2})$ When $x=\frac{\pi}{2}$: $y=\cos{\frac{\pi}{2}} = 0$, thus, the point is $(\frac{\pi}{2}, 0)$ When $x=\frac{3\pi}{4}$: Reference angle is $\frac{\pi}{4}$. Since the angle is in Quadrant II, cosine is negative. Thus, $y=\cos{\frac{3\pi}{4}} = -\cos{\frac{\pi}{4}}=-\frac{\sqrt2}{2}$ Thus, the point is $(\frac{3\pi}{4}, -\frac{\sqrt2}{2})$ When $x=\pi$: $y=\cos{\pi} = -1$, thus, the point is $(\pi, -1)$