sin2x-cos2x 怎么得到 根号2/2 sin(2x-pai/4)
热心网友
sin2x-cos2x 怎么得到 根号2/2 sin(2x-pai/4)将sin2x-cos2x乘以sin(π/4)=cos(π/4)=(√2)/2,再除以(√2)/2可得:[(√2)/2]*(sin2x-cos2x)=[sin2x*cos(π/4)-cos2x*sin(π/4)]=sin[2x-(π/4)]sin2x-cos2x=(√2)sin[2x-(π/4)]
热心网友
sin2x-cos2x =(根号2){sin2x×[1/(根号2)]-cos2x×[1/(根号2)]｝=(根号2){sin2x×[(根号2)/2]-cos2x×[(根号2)/2]｝=(根号2)(sin2x×cos45°-cos2x×sin45°)=(根号2)sin(2x-cos45°)=(根号2)sin[2x-(π/4)]有个公式啊：asinx+bcosx=[(a^2+b^2)^0.5]sin(x+B)其中，tgB=b/a另外，忘了告诉你：sin2x-cos2x≠根号2/2 sin(2x-π/4)它应该是：sin2x-cos2x＝(根号2)*sin(2x-π/4)