<s>[INST] Find the sum to n terms of the series Find the sum to n terms of the series
$$\frac {\sin x}{\cos x+\cos2x} + \frac {\sin2x}{\cos x+\cos4x} + 
\frac {\sin3x}{\cos x + \cos6x} +\dotsb $$
How can I solve this?
Here is what I did for the first term:
$$\frac {\sin\bigl(\frac{3x}{2} - \frac x2\bigr)}
  {2\cos\bigl(\frac{x}{2}\bigr)\cos\bigl(\frac {3x}{2}\bigr)}$$
After opening $N^r$ by $\sin(A-B) = \sin A\cos B - \cos A\sin B$ we get
$$ \frac 12\left(\tan\frac {3x}{2} - \tan\frac {x}{2}\right)$$
But this doesn't work out for the remaining terms. Am I missing something here or is this method completely wrong? Any hints? Help please!
edit : Please give a hint. I want to do it on my own :)
 [/INST] So I figured out the answer :
$ S = \frac {sinx}{cosx+cos2x} + \frac {sin2x}{cosx+cos4x}+ \frac {sin3x}{cosx+cos6x} + \quad ...$
Multiply both sides by $2sin\frac x2$
Take the first term :
$$\frac {2sin\frac x2sinx}{cosx+cos2x}$$
$$\frac {cos\frac{x}2-cos\frac {3x}2} {2cos\frac {3x}2cos\frac x2}$$
$$\frac 12(sec \frac {3x}2- sec \frac{x}2)$$
Now by same method we extend this to $n\ terms$ and add them
Finally we get,
$$2sin\frac x2S = \frac 12(sec (2n+1)\frac x2- sec \frac x2)$$
$$\therefore \quad S = \frac 14cosec\frac x2(sec(2n+1)\frac x2 - sec\frac x2)$$
This is the answer.
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