# heres the questionthe sum to n terms of the series(cos3x)/(sin2x)(sin4x) + (cos5x)/(sin4x)(sin6x) + (cos7x)/(sin6x)(sin8x)....isplz tell the steps to solve such a problemanswer works out to be1/2sinx{cosec2x - cosec(2n+2)x}thnx in advance..:-)

SAGAR SINGH - IIT DELHI
879 Points
11 years ago

Dear student,

Use the following series:

$\sin\left( x \right) \approx x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}.\!$

$\cos x = \sum^{\infin}_{n=0} \frac{(-1)^n}{(2n)!} x^{2n} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots\text{ for all } x\!$