两国集团划分世界:·[sin(2a+b)/sina]-2cos(a+b)=2,求sin^2b+2cos2a的值 怎么做的???

来源:百度文库 编辑:查人人中国名人网 时间:2024/04/30 08:53:48

原式可化为:[sin(2a+b)/sina]-2(cosa·cosb-sina·sinb)=2
sin(2a+b)-2sina·cosa·cosb+2sin^2a·sinb=2sina
sin2a·cosb+cos2a·sinb-2sina·cosa·cosb+2sin^2a·sinb=2sina
cos2a·sinb+2sin^2a·sinb=2sina
sinb(cos2a+2sin^2a)=2sina
sinb(cos^2a-sin^2a+2sin^2a)=2sina
sinb=2sina
sin^2b=4sin^2a
所求可化为:sin^2b+2-4sin^2a=2