Function: csc(x)=-2
Quadrant: 4
find cot.
Anyone? Last question of the year.
$\displaystyle
\frac{1}{sinx} = -2
$
$\displaystyle
-2sinx = 1
$
$\displaystyle
sinx = -\frac{1}{2}
$
$\displaystyle
sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}
$
easy to get $\displaystyle cotx$ from this
since $\displaystyle cosx(-\frac{\pi}{6}) = -\frac{\sqrt{3}}{2}$
and $\displaystyle cotx$ is $\displaystyle \frac{adj}{hyp} $ then $\displaystyle cotx$ is just $\displaystyle
-\sqrt{3}$
Hi BeSweet,
Last question? Wow. Have a very merry Christmas.
$\displaystyle \csc x = \frac{1}{\sin x}=-2$
$\displaystyle \sin x = -\frac{1}{2}$
Use $\displaystyle r^2=x^2+y^2$ to find x.
$\displaystyle \sin x =\frac{y}{r}=-\frac{1}{2}$
y = -1, r = 2
$\displaystyle 2^2=x^2+(-1)^2$
$\displaystyle x=\sqrt{3}$
$\displaystyle \tan x=\frac{y}{x}$
$\displaystyle \cot x = \frac{1}{\tan x}=\frac{x}{y}=\frac{\sqrt{3}}{-1}=-\sqrt{3}$