prove that (cos((nθ)/5)+isin((2nθ)/(10)))^5 −(1/e^(inθ) )=2sin(nθ) ?


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Question Number 103481 by mohammad17 last updated on 15/Jul/20

$p r o v e t h a t {(c o s \frac{n θ}{5} + i s i n \frac{2 n θ}{10})}^{5} - \frac{1}{e^{i n θ}} = 2 s i n (n θ) ?$
	Answered by Dwaipayan Shikari last updated on 15/Jul/20
	$e^(nθi) −e^(−inθ) =2sin(nθ) (e^(((nθ)/5)i) )^5 =e^(nθi) {cosnθ+isinnθ−cosnθ+isinnθ=2isinnθ$
	$e^{n θ i} - e^{- i n θ} = 2 s i n (n θ)$ ${(e^{\frac{n θ}{5} i})}^{5} = e^{n θ i} {c o s n θ + i s i n n θ - c o s n θ + i s i n n θ = 2 i s i n n θ$
	Commented by mohammad17 last updated on 15/Jul/20

	$t h a n k y o u s i r$
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