=[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
=2sinacosacosb+(cos²a-sin²a)sinb-2sinacosacosb+2sin²asinb
=cos²asinb-sin²asinb+2sin²asinb
=cos²asinb+sin²asinb
=(cos²a+sin²a)sinb
=sinb
∴sin(2a+b)/sina-2cos(a+b)=sinb/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)
如图
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sin(2a+b)\/sina - 2cos(a+b)
sin（2a+b）\/sina - 2cos（a+b）=[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
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这样，得到了积化和差的四个公式:sinαcosβ=[sin(α+β)+sin(α-β)]\/2 cosαsinβ=[sin(α+β)-sin(α-β)]\/2 cosαcosβ=[cos(α+β)+cos(α-β)]\/2 sinαsinβ=-[cos(α+β)-cos(α-β)]\/2 有了积化和差的四个公式以后,我们只需一个变形,就可以得到和差化积的四...
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sin(A+B)-sinA=2cos(A+B\/2)*sinB\/2怎么推导?
sin(A+B)-sinA=sin [(A+B\/2)+ B\/2]-sin[(A+B\/2)- B\/2]=[sin(A+B\/2)cos B\/2+cos(A+B\/2)sin B\/2]-[ sin(A+B\/2)cos B\/2-cos(A+B\/2)sin B\/2]=2 cos(A+B\/2)sin B\/2.∴原式成立.
