## Trigonometry (11th Edition) Clone

(a) $x = 2+sin~t$ $y = 1+cos~t$ We can see the graph below. (b) $(x-2)^2+(y-1)^2 = 1$
(a) $x = 2+sin~t$ $y = 1+cos~t$ When $t = 0$: $x = 2+sin~0 = 2$ $y = 1+cos~0 = 2$ When $t = \frac{\pi}{6}$: $x = 2+sin~\frac{\pi}{6} = 2.5$ $y = 1+cos~\frac{\pi}{6} = 1+\frac{\sqrt{3}}{2}$ When $t = \frac{\pi}{4}$: $x = 2+sin~\frac{\pi}{4} = 2+\frac{\sqrt{2}}{2}$ $y = 1+cos~\frac{\pi}{4} = 1+\frac{\sqrt{2}}{2}$ When $t = \frac{\pi}{3}$: $x = 2+sin~\frac{\pi}{3} = 2+\frac{\sqrt{3}}{2}$ $y = 1+cos~\frac{\pi}{3} = 1.5$ When $t = \frac{\pi}{2}$: $x = 2+sin~\frac{\pi}{2} = 3$ $y = 1+cos~\frac{\pi}{2} = 1$ When $t = \frac{2\pi}{3}$: $x = 2+sin~\frac{2\pi}{3} = 2+\frac{\sqrt{3}}{2}$ $y = 1+cos~\frac{2\pi}{3} = 0.5$ When $t = \pi$: $x = 2+sin~\pi = 2$ $y = 1+cos~\pi = 0$ When $t = \frac{4\pi}{3}$: $x = 2+sin~\frac{4\pi}{3} = 2-\frac{\sqrt{3}}{2}$ $y = 1+cos~\frac{4\pi}{3} = 0.5$ When $t = \frac{3\pi}{2}$: $x = 2+sin~\frac{3\pi}{2} = 1$ $y = 1+cos~\frac{3\pi}{2} = 1$ We can see the graph below. (b) $x = 2+sin~t$ $x-2 = sin~t$ $y = 1+cos~t$ $y-1 = cos~t$ $(x-2)^2+(y-1)^2 =sin^2~t+cos^2~t$ $(x-2)^2+(y-1)^2 = 1$