Finding the $\cot\left(\sin^{-1}\left(-\frac12\right)\right)$ How can I calculate this value?
$$\cot\left(\sin^{-1}\left(-\frac12\right)\right)$$
 A: Draw a right triangle (in the x>0,y<0 quadrant) with opposite edge -1 and hypotenuse 2.  Then the adjacent side is $\sqrt{2^2-1^2}=\sqrt{3}$.  cotangent is the ratio of adjacent side over opposite side.
A: You should probably have memorized things like the sine of 30 degrees. We therefore know that $sin(30) = 0.5$ So $arcsin(-1/2)=-30$ degrees
Now we want to take the cotangent of that. Well Cotangent is cosine over sine.
$cos(-30) = cos(30) = \sqrt(3)/2$
$sin(-30)=-sin(30)=-1/2$
Thus, the final answer is$-\sqrt(3)$
A: 
Arcsine is defined in the first and the fourth quadrants. If 
$\theta = \arcsin\left(-\dfrac 12 \right)$, then $\theta$ corresponds to the point
$(x,y)=(\sqrt 3, -1)$ with length $r=2$, because 
$\sin \theta = \dfrac yr = \dfrac{-1}{2}$, and the corresponding reference triangle shown above. Then $\cot\left(\arcsin\left(-\dfrac12\right)\right) = \dfrac xy = -\sqrt 3$.
A: $$\cot x =\frac{\cos x}{\sin x}=\frac{\sqrt{1-\sin^2x}}{\sin x}$$
 so we have $$\frac{\sqrt{\frac{3}{4}}}{-\frac{1}{2}}=\pm \sqrt{3}$$
There is not enough information in the problem to determine the sign.
A: Let arcsin(-1/2)=  x
Implies sinx=-1/2
Implies x= 120 degree
Now u need to find value of cot x
So  cot x= - underroot 3
