2*sin(60°-B)*sinB=cos(60°-2B)-cos60°_数学解答

2*sin(60°-B)*sinB=cos(60°-2B)-cos60°

人气：156 ℃　时间：2020-05-19 18:40:06

解答

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°