sin20^2+cos80^2+√3sin20cos80 别复制给我

根据cos2x=1-2(sinx)^2,
(sinx)^2=1/2(1-cos2x)
sin20^2+cos80^2+√3sin20cos80
=1/2(1-cos40°)+1/2(1+cos160°)+根号(3)/2(sin100°-sin60°)
=1-1/2cos40°-1/2cos20°-3/4+根号(3)/2sin100°
=1/4-1/2(cos40°+cos20°)+根号(3)/2sin80°
=1/4-cos10cos30°+根号(3)/2sin80°
=1/4-根号(3)/2cos10°+根号(3)/2sin80°
=1/4

sin20^2+cos80^2+√3sin20cos80
=sin20^2+sin10^2+√3sin20sin10
=sin20^2+sin10^2-2sin20sin10*cos150° (这儿相当于余弦定理)
=sin²150°
=1/4