# Problem with finding critical point of trig function

Can anyone help me out solving this? I don't understand how my professor ends up with $$1 - \frac{1}{2\sin^2(x/2)}$$ Thanks everyone!

The problem is stated in this picture: http://imgur.com/z8yw2

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What is the derivative of $\cot(x)$? –  N. S. Dec 17 '12 at 23:42
-csc^2(X) i believe? –  choloboy Dec 17 '12 at 23:43
Which is the same as $-\frac{1}{\sin^2(x)}$ ;) –  N. S. Dec 17 '12 at 23:56
Again, please type up your questions and don't just post a picture of them. –  Thomas Jan 31 '13 at 0:37
Using quotient rule to differentiate $\cot(x) = \cos(x)/\sin(x)$ gives
$$\cot'(x) = (-\sin(x)\sin(x) - \cos(x)\cos(x) )/\sin^2(x) = -1/\sin^2(x),$$ so if the argument is $x/2$ then by the chain rule we get $$\cot'(x/2) = -1/2\sin^2(x/2).$$