Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

2. Let $f(x)=x+\cot(x/2),\quad x\in\left[\frac\pi3,\frac{2\pi}3\right]$.

(a) Find all the critical numbers in the domain.

(b) Find the absolute maximum value and minimum value.

Solution: (a) $$f'(x)=1-\frac1{2\sin^2(x/2)}=0,\Rightarrow\sin(x/2)=\pm\frac1{\sqrt2}$$

It has only one solution $x=\frac\pi2$ in $\left[\frac\pi3,\frac{2\pi}3\right]$.

(b) Note that $f\left(\frac\pi2\right)=\frac\pi2+1,f\left(\frac\pi3\right)=\frac\pi3+\sqrt3,f\left(\frac{2\pi}3\right)=\frac{2\pi}3+1/\sqrt3$. Therefore,

Thus $f\left(\frac\pi3\right)=\frac\pi3+\sqrt3$ is the absolute maximum, $f\left(\frac\pi2\right)=\frac\pi2+1$ is the absolute minimum.

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

share|cite|improve this question
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

1 Answer 1

up vote 1 down vote accepted

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 $$ \cot'(x/2) = -1/2\sin^2(x/2). $$

share|cite|improve this answer
I edited your answer. Please use LaTeX (MathJax) to format questions and answers. You can find more about this here:… – Thomas Jan 31 '13 at 1:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.