## Trigonometry (11th Edition) Clone

$x = \frac{1}{3}~(tan^{-1}~2y - 2)$ where $y$ is in the interval $(-\infty, \infty)$
$y = \frac{1}{2}~tan~(3x+2)$ $x$ is restricted to the interval $(-\frac{2}{3}-\frac{\pi}{6}, -\frac{2}{3}+\frac{\pi}{6})$ We can find $y$ when $x$ approaches the value $~~-\frac{2}{3}-\frac{\pi}{6}~~$ from the right: $y = \frac{1}{2}~tan~(3x+2)$ $y = \frac{1}{2}~tan~[3( -\frac{2}{3}-\frac{\pi}{6})+2]$ $y = \frac{1}{2}~tan~[-2-\frac{\pi}{2}+2]$ $y = \frac{1}{2}~tan~(-\frac{\pi}{2})$ $y = -\infty$ We can find $y$ when $x$ approaches the value $~~-\frac{2}{3}+\frac{\pi}{6}~~$ from the left: $y = \frac{1}{2}~tan~(3x+2)$ $y = \frac{1}{2}~tan~[3( -\frac{2}{3}+\frac{\pi}{6})+2]$ $y = \frac{1}{2}~tan~[-2+\frac{\pi}{2}+2]$ $y = \frac{1}{2}~tan~(\frac{\pi}{2})$ $y = \infty$ When x is restricted to the interval $(-\frac{2}{3}-\frac{\pi}{6}, -\frac{2}{3}+\frac{\pi}{6})$, $y$ has a value on the interval $(-\infty, \infty)$ We can find an expression for $x$: $y = \frac{1}{2}~tan~(3x+2)$ $tan^{-1}~2y = 3x+2$ $x = \frac{1}{3}~(tan^{-1}~2y - 2)$ where $y$ is in the interval $(-\infty, \infty)$