Finding the $\cot\left(\sin^{-1}\left(-\frac12\right)\right)$

Question

How can I calculate this value?

$$\cot\left(\sin^{-1}\left(-\frac12\right)\right)$$

Answer 1

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$-\sqrt(3)$

Answer 2

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.

Answer 3

$$\cot x =\frac{\cos x}{\sin x}=\frac{\sqrt{1-\sin^2x}}{\sin x}$$ 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.

Answer 4

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