prove-that-cos-n-5-isin-2n-10-5-1-e-in-2sin-n-

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^{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

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