sin2B+sin2C=2sin(B+C)cos(B-C)
问题描述:
sin2B+sin2C=2sin(B+C)cos(B-C)
怎样证明,帮我推出来谢谢
答
左边=sin[(B+C)+(B-C)]+sin[(B+C)-(B-C)]
=sin(B+C)cos(B-C)+cos(B+C)sin(B-C)+sin(B+C)cos(B-C)-cos(B+C)sin(B-C)
=2sin(B+C)cos(B-C)