=(sin20°)^2+(sin40°)^2-2sin20°*sin40°*cos120° =(sin120°)^2=3/4.

问题描述:

=(sin20°)^2+(sin40°)^2-2sin20°*sin40°*cos120° =(sin120°)^2=3/4.

构造△ABC,A=20°,B=40°,C=120°,
由余弦定理得a^2+b^2-2abcosC=c^2,
利用正弦定理得=(sin°A)^2+(sinB)^2-2sin*AsinB*cosC =(sinC)^2,
即(sin20°)^2+(sin40°)^2-2sin20°*sin40°*cos120° =(sin120°)^2