第1个回答 2019-06-21
consider
(cosx +isinx)^3= [(cosx)^3 -3(cosx)(sinx)^2] +i[cosx.(sinx)^3 -3(sinx)^3 ]
cos3x +isin3x = [(cosx)^3 -3(cosx)(sinx)^2] +i[cosx.(sinx)^3 -3(sinx)^3 ]
=>
cos3x
= (cosx)^3 -3(cosx)(sinx)^2
= (cosx)^3 -3(cosx)[1-(cosx)^2]
=4(cosx)^3 - 3cosx
4(cosx)^3 = cos3x +3cosx
(cosx)^3 = (1/4) [cos3x +3cosx]
x=3t
(cos3t)^3 = (1/4) [cos9t +3cos3t]本回答被网友采纳