Question

if 1+sinΦ+sin^2Φ+.........+infinate =4+2√3​

Answer 1

Answer:

60 or π/3 in (-π,π)

Step-by-step explanation:

Series is GP

with

first term(a)=1

and

common ratio (r) =sinα

Now,

sum of infinite GP is given by

$= \frac{a}{1 - r} \\ = \frac{1}{1 - \sin( \alpha ) } = 4 + 2 \sqrt{3}$

cross multiply

$\frac{1}{4 + 2 \sqrt{3} } = 1 - \sin( \alpha ) \\ rationalize \\ \frac{4 - 2 \sqrt{3} }{16 - 12} = 1 - \sin( \alpha ) \\ = \frac{4}{4} - \frac{2 \sqrt{3} }{4} = 1 - \sin( \alpha ) \\ - \frac{2 \sqrt{3} }{4} = - \sin( \alpha ) \\ \frac{ \sqrt{3} }{2} = \sin( \alpha ) \\ hence \\ \alpha = 60 \: deg \: or \: \frac{\pi}{3} \: in \: ( - \pi \: to \: \pi)$

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