设角A=-35/6π则{2sin(π+a)cos(π-a)-cocs(π+a)}/1+sin^2+sin(π-a)cos^2(π+a)为

问题描述:

设角A=-35/6π则{2sin(π+a)cos(π-a)-cocs(π+a)}/1+sin^2+sin(π-a)cos^2(π+a)为

sina=sin(-35π/6)=sin(-6π+π/6)=1/2 cosα=根号3/2
2sin(π+α)cos(π-α)-cos(π+α)/( 1+sin^2α)+sin(π-α)-cos^2(π+α)
=2(-sinα)(-cosα)+cosα/(1+sin²α)+sinα-cos²α=根号3/2+(根号3/2)*4/5+1/2-3/4=9根号3/10-1/4