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Question-171979




Question Number 171979 by Shrinava last updated on 22/Jun/22
Answered by puissant last updated on 23/Jun/22
sinθ+icosθ=((−i)/(−i))•sinθ+icosθ                             = ((−isinθ+cosθ)/(−i))                             = −(e^(−iθ) /i)
$${sin}\theta+{icos}\theta=\frac{−{i}}{−{i}}\bullet{sin}\theta+{icos}\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{−{isin}\theta+{cos}\theta}{−{i}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:−\frac{{e}^{−{i}\theta} }{{i}} \\ $$

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