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
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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)=sinb/sina
左边=[sin(2a+b)-2cos(a+b)sina]/sina
=[sina*cos(a+b)+sin(a+b)*cosa-2cos(a+b)sina]/sina
=[sin(a+b)*cosa-cos(a+b)sina]/sina
=sin(a+b-a)/sina
=sinb/sina
=右边
所以结论成立.
逆推法：
sin(2a+b)/sina-2cos(a+b)=sinb/sina
<==>sin(a+a+b)-2cos(a+b)sina=sinb
<==>sinacos(a+b)+cosasin(a+b)-2cos(a+b)sina=sinb
<==>cosasin(a+b)-cos(a+b)sina=sinb
<==>sin(a+b-a)=sinb
<==>sinb=sinb恒成立
以上各步可逆
证毕

左=(sin(2a+b)-2cos(a+b)sina)/sina
=(sin(2a+b)-(sin(a+b+a)-sin(a+b-a)))/sina
=sinb/sina
运用了三角函数的积化和差
2cosasinb=sin(a+b)-sin(a-b)

∵sin(2a+b)-2sinacos(a+b)=sin2acosb+cos2asinb-2sina(cosacosb-sinasinb)
=2sinacosacosb+(cos²a-sin²a)sinb-2sinacosacosb+2sin²asinb
=cos²asinb-sin²asinb+2sin²asinb
=cos²asinb+sin²asinb
=(cos²a+sin²a)sinb
=sinb
∴sin(2a+b)/sina-2cos(a+b)=sinb/sina

如图

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