1-sin2x-1-sin2-2-x-1-sin2-n-x-Find-the-value-

Question Number 104122 by Dwaipayan Shikari last updated on 19/Jul/20

\frac{1}{\sin 2 x} + \frac{1}{{\sin 2}^{2} x} + \dots . . + \frac{1}{{\sin 2}^{n} x}

Find the value

Answered by OlafThorendsen last updated on 19/Jul/20

\frac{1}{\sin u} = \frac{\sin (u - \frac{u}{2})}{s i n u \sin \frac{u}{2}}

\frac{1}{\sin u} = \frac{\sin u \cos \frac{u}{2} - \sin \frac{u}{2} \cos u}{s i n u \sin \frac{u}{2}}

\frac{1}{\sin u} = \frac{\cos \frac{u}{2}}{\sin \frac{u}{2}} - \frac{\cos u}{\sin u}

\frac{1}{\sin u} = \cot \frac{u}{2} - \cot u

u = 2^{k} x

\frac{1}{{\sin 2}^{k} x} = {\cot 2}^{k - 1} x - {\cot 2}^{k} x

\underset{k = 1}{\sum^{n}} \frac{1}{{\sin 2}^{k} x} = \underset{k = 1}{\sum^{n}} {\cot 2}^{k - 1} x - {\cot 2}^{k} x

= \cot x - \cot 2 x

+ \cot 2 x - \cot 4 x

+ \dots

+ {\cot 2}^{n - 1} x - {\cos 2}^{n} x

\underset{k = 1}{\sum^{n}} \frac{1}{{\sin 2}^{k} x} = \cot x - {\cot 2}^{n} x

Commented by Dwaipayan Shikari last updated on 19/Jul/20

Great sir!