((sin1)/e)−((sin(2))/(2e^2 ))+((sin(3))/(3e^3 ))−((sin(4))/(4e^4 ))+...=tan^(−1) (((sin(1))/(cos(1)+e)))


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Question Number 133692 by Dwaipayan Shikari last updated on 23/Feb/21

$\frac{s i n 1}{e} - \frac{s i n (2)}{2 e^{2}} + \frac{s i n (3)}{3 e^{3}} - \frac{s i n (4)}{4 e^{4}} + . . . = {t a n}^{- 1} (\frac{s i n (1)}{c o s (1) + e})$
Commented by Dwaipayan Shikari last updated on 23/Feb/21

$S h a r e a n I d e a t o s o l v e t h i s . I h a v e a s t r a n g e w a y b u t i w a n t$ $t o s e e {o t h e r}^{'} s p e r s p e c t i v e ?$
	Answered by mindispower last updated on 23/Feb/21

	$= \frac{s i n (k)}{{n e}^{k}} = I m \sum_{m ⩾ 1} {(- 1)}^{m} \frac{e^{i m}}{{m e}^{m}}$ $= I m - l n (1 + e^{(i - 1)})$ $= - a r g (1 + e^{i} . e^{- 1}) = - {t g}^{-} (\frac{e^{- 1} s i n (1)}{1 + e^{- 1} c o s (1)})$ $= - {t g}^{-} (\frac{s i n (1)}{e + c o s (1)}) = \frac{π}{2} - {t g}^{-} (\frac{e + c o s (1)}{s i n (1)})$
	Commented by Dwaipayan Shikari last updated on 23/Feb/21

	$S i r f i r s t l i n e \underset{m ⩾ 1}{\sum^{\infty}} \frac{{(- 1)}^{m + 1}}{m e} e^{i m}$
	Commented by mindispower last updated on 23/Feb/21

	$t h a n x y e s s i r s o r r y t o o b u s y t h i s d a y s$ $b a c k s o o n$
	Answered by Dwaipayan Shikari last updated on 23/Feb/21

	$l o g (1 + {a e}^{i θ}) = l o g (\sqrt{2 + 2 a c o s θ}) + {i t a n}^{- 1} \frac{a s i n θ}{a c o s θ + 1}$ ${a e}^{i θ} - \frac{a^{2} e^{2 i θ}}{2} + \frac{a^{3} e^{3 i θ}}{3} - . . . = l o g (\sqrt{2 + 2 a c o s θ}) + {i t a n}^{- 1} \frac{a s i n θ}{a c o s θ + 1}$ $a c o s θ - \frac{a^{2}}{2} c o s 2 θ + \frac{a^{3}}{3} c o s 3 θ - . . . = l o g (\sqrt{2 + 2 a c o s θ})$ $a s i n θ - \frac{a^{2}}{2} s i n 2 θ + \frac{a^{3}}{3} s i n 3 θ - . . . = {t a n}^{- 1} \frac{a s i n θ}{a c o s θ + 1}$ $a = \frac{1}{e}$
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