Question

在△ABC中,角C=30°,则sin²A+sin²B-2sinB*sinA*cosC的值是?

Answer 1

应该是4(sinC)^2=1

设AB边上过C点的高为h­a

sin²A+sin²B-2sinB*sinA*cosC

=(ha/b)²+(ha/a)²-2(ha²/ab)cosC

=ha²/a²b²(a²+b²-2ab*cosC)

=(ha²/a²b²)(c²)

=(c*ha/ab)²

c*ha=2S三角形ABC

=(2S/ab)²

S/a为三角形在BC边上的高,再除以b为sin30°

=4sin²30°

=1