化简sin(2x-pi)sin[x+(3/2)pi]/cos[x+(5/2)pi][1-sin(pi+x)]
问题描述:
化简sin(2x-pi)sin[x+(3/2)pi]/cos[x+(5/2)pi][1-sin(pi+x)]
答
sin(2x-pi)sin[x+(3/2)pi]/cos[x+(5/2)pi][1-sin(pi+x)]=(-sin2x)(cosx)/(-sinx)(1+sinx)=(2sinxcos²x)/(sinx)(1+sinx)=(2cos²x)/(1+sinx)=(1+cos2x)/(1+sinx)