## Trigonometry (11th Edition) Clone

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