ldcf.net
ǰλãҳ >> ΪʲôCosx+Cox3x=2Cos2x*Cosx >>

ΪʲôCosx+Cox3x=2Cos2x*Cosx

cos2x-x+cos2x+x =cos2xcosx+sin2xsinx+cos2xcosx-sin2xsinx =2cos2xcosx

e^(ix)=cosx+isinx e^(ix)+e^(i2x)+e^(i3x)++e*(inx)=(cosx+cos2x++cosnx)+i(sinx+sin2x++sinnx) =[e^(inx+ix) -e^(ix)]/[e^(ix)-1] һȺҶ˷ֳʵ鲿(ĸͷͬ (cosx-1)-isinxȺʵ鲿...

ɴ޷

cosxcos2xcos3xdx 1/2ңcosxcos3xcos3xdx1/2cosxcos3xdx1/2ңcos3x^2dx 1/4ңcos2xcos4xdx1/4ң1cos6xdx 1/4dx1/4cos2xdx1/4cos4xdx1/4cos6xdx 1/4x1...

Ҧ/2 0 (cos2x/cosx+sinx)dx =Ҧ/2 0 (cos²x-sin²x)/(cosx+sinx)dx =Ҧ/2 0 (cosx-sinx)dx =sinx+cosx /2 0 =(1+0)-(0+1) =0

վҳ | վͼ
All rights reserved Powered by www.ldcf.net
copyright ©right 2010-2021
磬ַϵͷzhit325@qq.com