Flag Trigonometry> solve this plzzz (trig)...
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heres the question

the sum to n terms of the series

(cos3x)/(sin2x)(sin4x) + (cos5x)/(sin4x)(sin6x) + (cos7x)/(sin6x)(sin8x)....is

plz tell the steps to solve such a problem

answer works out to be

1/2sinx{cosec2x - cosec(2n+2)x}

thnx in advance..:-)

arpan chandrashekhar , 14 Years ago
Grade
anser 1 Answers
SAGAR SINGH - IIT DELHI

Last Activity: 14 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\!

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