Taylor Series of $\frac{1}{1-\cos x}$ The problem is, as the title suggests, to find the Power Series Expansion of $\frac{1}{1- \cos x}$ around $x=c$.

What I've tried:


Direct Computation: Derivatives get very ugly quickly, and don't yield a nice formula that I can recognize as a "series."
Tried finding the integral of $\frac{1}{1- \cos x}$, finding it's series and then differentiating it to get the new series.
Tried reverse of the above, differentiating and finding it's series, then integrating (very messy).
Then I tried some substitution "tricks", like using the series of $\frac{1}{1-x}$ and then plugging in the series expansion for $\cos x$, but that's a double sum that I struggled to produce anything useful from
:$\displaystyle \sum_{k=0}^\infty\left(\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}\right)^k$


I am literally at my witts end with this problem. I have spent perhaps a day or two trying to figure it out, because I feel that I am so close - but just barely missing something. I do not want the solution posted - now it's personal and I have to figure it out, but I would greatly appreciate a hint in the right direction, or to point out a mistake that I may be overlooking.
As @hjpotter92 suggest, you have 
$$\frac{1}{1-\cos(x)} = \frac{1}{2\sin^2(x/2)} = -\frac{\text{d}}{\text{d}x}(\cot(x/2)).$$
Now you can exploit the series expansion of 
$$\cot(x)=\sum_{n=0}^\infty\frac{(-1)^n 2^{2n}B_{2n}}{(2n)!}x^{2n-1}, \quad \forall 0&lt;\left|x\right|&lt;\pi.$$ 
Now, by evaluating in $x/2$, differentiating each coefficient and changing the sign, you get
$$\frac{1}{1-\cos(x)} =\sum_{n=0}^\infty\frac{(-1)^{n+1} 2(2n-1)B_{2n}}{(2n)!}x^{2n-2}, \quad \forall 0&lt;\left|x\right|&lt;2\pi,$$
where $B_n$ are the Bernoulli numbers.
integration of 1/1-cos(x)=
= 1/1-cos(x)(1+cos(x)/1+cos(x) 
= 1+cos(x)/[1-cos(x)][1+cos(x)] 
= 1+cos(x)/[sin(x)][sin(x) 
= 1/sin(x)[sin(x)]+cos(x)/sin(x)[sin(x)] 
= cosec(x)[cosec(x)+cot(x)cosec(x) 
= -cot(x)+cosec(x)+c