## Trigonometry (11th Edition) Clone

(a) $x = 4~sin~t$ $y = 3~cos~t$ We can see the graph below. (b) $\frac{x^2}{16}+\frac{y^2}{9} = 1$
(a) $x = 4~sin~t$ $y = 3~cos~t$ When $t = 0$: $x = 4~sin~0 = 0$ $y = 3~cos~0 = 3$ When $t = \frac{\pi}{6}$: $x = 4~sin~\frac{\pi}{6} = 2$ $y = 3~cos~\frac{\pi}{6} = \frac{3\sqrt{3}}{2}$ When $t = \frac{\pi}{4}$: $x = 4~sin~\frac{\pi}{4} = 2\sqrt{2}$ $y = 3~cos~\frac{\pi}{4} = \frac{3\sqrt{2}}{2}$ When $t = \frac{\pi}{3}$: $x = 4~sin~\frac{\pi}{3} = 2\sqrt{3}$ $y = 3~cos~\frac{\pi}{3} = \frac{3}{2}$ When $t = \frac{\pi}{2}$: $x = 4~sin~\frac{\pi}{2} = 4$ $y = 3~cos~\frac{\pi}{2} = 0$ When $t = \frac{2\pi}{3}$: $x = 4~sin~\frac{2\pi}{3} = 2\sqrt{3}$ $y = 3~cos~\frac{2\pi}{3} = -\frac{3}{2}$ When $t = \pi$: $x = 4~sin~\pi = 0$ $y = 3~cos~\pi = -3$ When $t = \frac{4\pi}{3}$: $x = 4~sin~\frac{4\pi}{3} = -2\sqrt{3}$ $y = 3~cos~\frac{4\pi}{3} = -\frac{3}{2}$ When $t = \frac{3\pi}{2}$: $x = 4~sin~\frac{3\pi}{2} = -4$ $y = 3~cos~\frac{3\pi}{2} = 0$ We can see the graph below. (b) $x = 4~sin~t$ $y = 3~cos~t$ $\frac{x^2}{16}+\frac{y^2}{9} = \frac{(4~sin~t)^2}{16}+\frac{(3~cos~t)^2}{9}$ $\frac{x^2}{16}+\frac{y^2}{9} = \frac{16~sin^2~t}{16}+\frac{9~cos^2~t}{9}$ $\frac{x^2}{16}+\frac{y^2}{9} = sin^2~t+cos^2~t$ $\frac{x^2}{16}+\frac{y^2}{9} = 1$