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How can I calculate this value?

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

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    $\begingroup$ Draw a triangle. $\endgroup$ – user61527 Jul 13 '14 at 1:19
  • $\begingroup$ The opposite side to the angle is allowed to be negative? $\endgroup$ – user8028 Jul 13 '14 at 1:20
  • $\begingroup$ Sure, the triangle just points downwards instead of upwards. Alternatively, the fact that $\sin$ and $\cot$ are both odd means that $\cot(\arcsin(-1/2)) = - \cot(\arcsin(1/2))$. $\endgroup$ – user61527 Jul 13 '14 at 1:21
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    $\begingroup$ Holy oversized cotangent, Batman. $\endgroup$ – user61527 Jul 13 '14 at 1:22
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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$-\sqrt(3)$

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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.

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

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  • $\begingroup$ You need to include why $\cos x$ is $\ge0,$ here? $\endgroup$ – lab bhattacharjee Jul 13 '14 at 4:59
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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 $\cot\left(\arcsin\left(-\dfrac12\right)\right) = \dfrac xy = -\sqrt 3$.

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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

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    $\begingroup$ See help on formatting and please edit the post accordingly. $\endgroup$ – user147263 Jul 13 '14 at 3:22

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