证明sin(2a+b)/sina-2cos(a+b)=sinb/sina

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证明sin(2a+b)/sina-2cos(a+b)=sinb/sina证明sin(2a+b)/sina-2cos(a+b)=sinb/sina证明sin(2a+b)/sina-2cos(a+b)=s

证明sin(2a+b)/sina-2cos(a+b)=sinb/sina
证明sin(2a+b)/sina-2cos(a+b)=sinb/sina

证明sin(2a+b)/sina-2cos(a+b)=sinb/sina
sin(2a+b)=sin2acosb+cos2asinb=2sinacosacosb+(1-2sin²a)sinb
2cos(a+b)=2cosacosb-2sinasinb
代入原式得：2cosacosb+sinb/sina-2sinasinb-2cosacosb+2sinasinb=sinb/sina
即：sin(2a+b)/sina-2cos(a+b)=sinb/sina
就是把sin(2a+b)和2cos(a+b)展开,代入等式左边即可

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题误！原题等价于证明：[sin(2a+b)-sinb]/sinb-2cos(a+b)=[sinb-sina]/sinb,而sin(2a+b)-sinb=2sinacos(a+b),所以只要证明2cos(a+b)[sina/sinb-1]=(sinb-sina)/sina……①，若a=b，①式显然成立，若a≠b，则只需证明2cos(a+b)/sinb=-1/sina,即需证明2cos(a+b)sina+sinb=0,而sinb=sin(a+b-a)=sin(a+b)cosa-sinacos(a+b),所以要证sin(a+b)cosa+sinacos(a+b)=sin(2a+b)=0成立，而题设没有这一条件！事实上令a=-b,原等式显然不成立。

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只要把sin(2a+b)按公式展开再合并就可以了，详细如下
2a+b=(a+b)+a
左边=sin(a+b)cosa/sina+cos(a+b)sina/sina-2cos(a+b)
=sin(a+b)cosa/sina-cos(a+b)
=sin(a+b)cosa/sina-cos(a+b)sina/sina
=sin(a+b-a)/sina
=右边

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