## Trigonometry (11th Edition) Clone

$\frac{\sqrt3}{2}$
RECALL: (1) $\theta = \\csc^{-1}{(x)} \longrightarrow \csc{\theta}=x$ (2) $y=\csc^{-1}{x}$ is defined only in $[-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}]$, which means the angle is either in Quadrant I or IV. (3) $\csc{\theta} = \frac{r}{y}$ (4) $\cos{\theta} = \frac{x}{r}$, where $r^2=x^2+y^2$ Let $\theta = \csc^{-1}{(-2)}$ This means that $\csc{\theta} = -2$. Since cosecant is negative, then the angle is in Quadrant IV. With $\csc{\theta} = -2=\frac{2}{-1}=\frac{r}{y}$, we can set $y=-1$ and $r=2$. Solve for $x$ using the formula $r^2=x^2+y^2$ to obtain: $r^2=x^2+y^2 \\2^2=x^2+(-1)^2 \\4=x^2+1 \\4-1=x^2 \\3=x^2 \\\pm\sqrt3=x$ Since the angle is in Quadrant IV, then $x$ is positive so $x=\sqrt3$. Thus, $\cos{(\csc^{-1}{(-2)})} \\=\cos{\theta} \\= \frac{x}{r} \\= \frac{\sqrt3}{2}$