三角形ABC中,角C=30度,求(sinA)^2+(sinB)^2-2sinAsinBcosC的值
问题描述：
三角形ABC中,角C=30度,求(sinA)^2+(sinB)^2-2sinAsinBcosC的值
答
(sinA)^2+(sinB)^2-2sinAsinBcosC
= [sin(B+30°)]^2+(sinB)^2-2sin(B+30°)sinBcos30°
= 3/4(sinB)^2+√3/2sinBcosB+1/4(cosB)^2+(sinB)^2-3/2(sinB)^2-√3/2sinBcosB
= 1/4
(注:sinA=sin(180°-B-C)=sin(B+C)=sin(B+30°)