2*sin(60°-B)*sinB=cos(60°-2B)-cos60°
问题描述:
2*sin(60°-B)*sinB=cos(60°-2B)-cos60°
答
2*sin(60°-B)*sinB
=2*(sin60°*cosB-cos60°*sinB)*sinB
=sin60°*2*cosB*sinB-cos60°*2*sinB*sinB
=sin60°*sin2B-cos60°(1-cos2B)
=cos60°*cos2B+sin60°*sin2B-cos60°
=cos(60°-2B)-cos60°