+0  
 
0
47
2
avatar+44 

Prove that  \(sin\theta =\frac{e^{i\theta }-e^{-i\theta }}{2i}\)

 Jul 16, 2021
 #2
avatar+114112 
+1

like this

 

\(sin\theta =\frac{e^{i\theta}-e^{-i\theta}}{2i}\\~\\ RHS=\frac{[cos\theta+isin\theta]-[cos(-\theta)+isin(-\theta)]}{2i}\\ RHS=\frac{[cos\theta+isin\theta]-[cos(\theta)-isin(\theta)]}{2i}\\ RHS=\frac{2isin\theta}{2i}\\ RHS=sin\theta\\~\\ RHS=LHS \qquad \qquad Q.E.D.\)

 Jul 17, 2021

17 Online Users

avatar