根号2/2cos(2x+π/4)+sin^2 x 求化简过程根号2/2cos(2x+π/4)+sin^2 x=1/2cos2x-1/2sin2x+1/2(1-cos2x)=1/2-1/2sin2x 求化简过程,谢谢
问题描述:
根号2/2cos(2x+π/4)+sin^2 x 求化简过程
根号2/2cos(2x+π/4)+sin^2 x=1/2cos2x-1/2sin2x+1/2(1-cos2x)=1/2-1/2sin2x 求化简过程,谢谢
答
cos(2x+π/4)=cosπ/4cos2x-sinπ/4sin2x
cosπ/4=sinπ/4=2分之根号2
cos2x=cosx^2-sinx^2=1-2sinx^2
sinx^2=(1-cos2x)/2