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

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

$${prove}\:{that}\:\left({cos}\frac{{n}\theta}{\mathrm{5}}+{isin}\frac{\mathrm{2}{n}\theta}{\mathrm{10}}\right)^{\mathrm{5}} −\frac{\mathrm{1}}{{e}^{{in}\theta} }=\mathrm{2}{sin}\left({n}\theta\right)\:? \\ $$

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

$${e}^{{n}\theta{i}} −{e}^{−{in}\theta} =\mathrm{2}{sin}\left({n}\theta\right) \\ $$$$\left({e}^{\frac{{n}\theta}{\mathrm{5}}{i}} \right)^{\mathrm{5}} ={e}^{{n}\theta{i}} \:\:\:\:\:\left\{{cosn}\theta+{isinn}\theta−{cosn}\theta+{isinn}\theta=\mathrm{2}{isinn}\theta\right. \\ $$

Commented by mohammad17 last updated on 15/Jul/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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