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Question Number 134475 by EDWIN88 last updated on 04/Mar/21

If 1 + \cos x + \cos 2 x + \cos 3 x + \cos 4 x + \dots + \infty = 3

for 0 < x ⩽ \frac{π}{2}

find the value of \sin x + \sin 2 x + \sin 3 x + \dots + \infty

Commented by Dwaipayan Shikari last updated on 04/Mar/21

\underset{n = 0}{\sum^{\infty}} c o s (n x) = \frac{1}{2} \underset{n = 0}{\sum^{\infty}} e^{i n x} + \frac{1}{2} \underset{n = 0}{\sum^{\infty}} e^{- i n x} = \frac{1}{2} . \frac{1}{1 - e^{i x}} + \frac{1}{2} . \frac{e^{i x}}{e^{i x} - 1}

= \frac{1}{2}

\underset{n = 1}{\sum^{\infty}} \frac{c o s (n x)}{n} = - l o g (4 {s i n}^{2} \frac{x}{2})

- \underset{n = 1}{\sum^{\infty}} s i n (n x) = - \frac{d}{d x} l o g (4 {s i n}^{2} \frac{x}{2})

\underset{n = 1}{\sum^{\infty}} s i n (n x) = \frac{2 s i n x}{2 - 2 c o s x} = \frac{s i n x}{1 - c o s x}

Commented by benjo_mathlover last updated on 04/Mar/21

what the answer sir ?

Commented by Dwaipayan Shikari last updated on 04/Mar/21

I f

1 + c o s x + {c o s}^{2} x + {c o s}^{3} x + \dots = \frac{1}{1 - c o s x} = 3

\Rightarrow c o s x = \frac{2}{3} \Rightarrow s i n x = \frac{\sqrt{5}}{3}

T h e n s i n x + {s i n}^{2} x + \dots = \frac{s i n x}{1 - s i n x} = \frac{\sqrt{5}}{3 - \sqrt{5}} = \frac{\sqrt{5}}{4} (\sqrt{5} + 3)

= \frac{1}{4} (3 \sqrt{5} + 5) :)

Commented by EDWIN88 last updated on 04/Mar/21

The available answer

(a) 2 (\sqrt{5} - 3) (c) \frac{1}{4} (3 \sqrt{5} - 5)

(b) \frac{1}{2} (2 \sqrt{3} + 2) (d) \frac{1}{4} (3 \sqrt{5} + 5)

(e) \frac{1}{2} (2 \sqrt{3} - 2)

Commented by Dwaipayan Shikari last updated on 04/Mar/21

M a y b e e r r o r i n y o u r q u e s t i o n s i r!

Commented by EDWIN88 last updated on 04/Mar/21

yes sir . i think it wrong

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