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Question Number 104243 by M±th+et+s last updated on 20/Jul/20

p r o v e t h a t

π = 2 \times \frac{2}{\sqrt{2}} \times \frac{2}{\sqrt{2 + \sqrt{2}}} \times \frac{2}{\sqrt{2 + \sqrt{2 + \sqrt{2}}}} \times \dots . .

Answered by OlafThorendsen last updated on 20/Jul/20

\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}

\sin x = 2^{2} \sin \frac{x}{4} \cos \frac{x}{4} \cos \frac{x}{2}

\dots

\sin x = 2^{n} \sin \frac{x}{2^{n}} \cos \frac{x}{2^{n}} \cos \frac{x}{2^{n - 1}} \dots

to be continued \dots

Answered by Dwaipayan Shikari last updated on 20/Jul/20

2 . \frac{2}{\sqrt{2}} . \frac{2}{\sqrt{2 + \sqrt{2}}} . \frac{2}{\sqrt{2 + \sqrt{2 + \sqrt{2}}}} \dots \dots

\lim_{n \to \infty} S_{n} = 2^{n} (\frac{1}{\sqrt{2}} . \frac{1}{\sqrt{2 + \sqrt{2}}} . \frac{1}{\sqrt{2 + \sqrt{2 + \sqrt{2}}}} \dots \dots n)

S_{n} = 2^{n} \cos \frac{π}{4} . \frac{1}{2 \cos \frac{π}{8} 2} . \frac{1}{\cos \frac{π}{16} 2} . \frac{1}{\cos \frac{π}{32}} \dots . . n

S_{n} = 2^{n} \cos \frac{π}{4} . \frac{1}{2^{n}} (\frac{1}{\cos (0) \dots \dots . . \cos \frac{π}{8}})

S_{n} = \cos \frac{π}{4} . (\frac{1}{\cos \frac{π}{8}} . \frac{1}{\cos \frac{π}{16}} . \frac{1}{\cos \frac{π}{32}} \dots .) continue