(COS + iSIN )N =COSN + iSIN N. PROOF:- Consider, (Cos1+ iSin 1 )(Cos + iSin ) = (cos 1 cos - Sin 1Sin ) + i ( Sin 1 Cos + Sin Cos1 ) = Cos ( 1 + ) + i Sin (1 + ) Hence, (Cos1 + iSin 1 )(Cos + iSin ) = Cos ( 1 + ) + i Sin (1 + )--eq1 Consider = 2 + 3------eq2 From the above eq2 , (Cos + iSin ) = Cos ( 2 + 3 ) + i Sin (2 + 3 )-----------eq3 From proof eq1, we have Cos ( 2 + 3 ) + i Sin (2 + 3) = (Cos2 + iSin 2 )(Cos3 + iSin 3 )------eq4 From eq3 and eq4, (Cos + iSin ) = (Cos2 + iSin 2 )(Cos3 + iSin 3 )------eq5 From eq1,eq5, (Cos1 + iSin 1 )(Cos2 + iSin 2 )(Cos3 + iSin 3 ) =Cos ( 1 + 2 + 3) + i Sin (1 + 2 + 3 )----eq6 Now,from step2 it is again repeated This goes on......................... Consider all ' s to be equal.......... u will get, (Cos + iSin )(Cos + iSin ).......(Cos + iSin ) =Cos ( + +.... + ) + i Sin ( + +....+ ) i.e. (COS + iSIN )N =COSN + iSIN N. CHEERS !
|
Moderation Team
![[Post New]](/templates/default/images/icon_minipost_new.gif){kind=icon}
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
216
