Prove-that-tanx-2-pi-2x-2-pi-2x-2-3x-2pi-2-3x-2pi-2-5x-2pi-2-5x-2pi-

Question Number 122740 by Dwaipayan Shikari last updated on 19/Nov/20

P r o v e t h a t

t a n x = \frac{2}{π - 2 x} - \frac{2}{π + 2 x} + \frac{2}{3 x - 2 π} - \frac{2}{3 x + 2 π} + \frac{2}{5 x - 2 π} - \frac{2}{5 x + 2 π} + \dots .

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

s i n x h a s v a l u e 0 a t 0, π, - π, 2 π, - 2 π, \dots .

S o i t c a n b e w r i t t e n a s p r o d u c t s

s i n x = C x (π - x) (x + π) (2 π - x) (x + 2 π) \dots

x \to 0 \frac{s i n x}{C x} = (π - x) (x + π) (2 π - x) \dots

\Rightarrow \frac{1}{C} = π . π . 2 π . 2 π \dots

S o s i n x = x (1 - \frac{x}{π}) (1 + \frac{x}{π}) (1 - \frac{x}{2 π}) \dots

\frac{s i n x}{x} = \underset{n = 1}{\prod^{\infty}} (1 - \frac{x^{2}}{n^{2} π^{2}})

S o i n t h i s m a n n e r

c o s x = (1 - \frac{2 x}{π}) (1 + \frac{2 x}{π}) (1 - \frac{2 x}{3 π}) (1 + \frac{2 x}{3 π}) \dots

l o g (c o s x) = l o g (1 - \frac{2 x}{π}) + l o g (1 + \frac{2 x}{π}) + \dots

- \frac{s i n x}{c o s x} = \frac{\frac{- 2}{π}}{1 - \frac{2 x}{π}} + \frac{\frac{2}{π}}{1 + \frac{2 x}{π}} - \frac{\frac{2}{3 π}}{1 - \frac{2 x}{3 π}} + \dots

t a n x = \frac{2}{π - 2 x} - \frac{2}{π + 2 x} + \frac{2}{3 π - 2 x} - \frac{2}{3 π + 2 x} + \frac{2}{5 π - 2 x} - \frac{2}{5 π + 2 x} + \dots

t a n x = \frac{1}{\frac{π}{2} - x} - \frac{1}{\frac{π}{2} + x} + \frac{1}{\frac{3 π}{2} - x} - \frac{1}{\frac{3 π}{2} + x} + \frac{1}{\frac{5 π}{2} - x} - \frac{1}{\frac{5 π}{2} + x} + \dots

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

I {d o n}^{'} t h a v e t h e j u s t i f i c a t i o n f o r t h i s p r o o f . I s t h i s r i g h t s i r ?

Commented by mindispower last updated on 19/Nov/20

c o s (x) = s i n (\frac{π}{2} - x)

\frac{d}{d x} l n c o s (x) = - t g (x)