sin(2x+y)=3siny,x≠kπ+π/2,x+y≠kπ+π/2(k∈Z)求证:tan(x+y)=2tanx
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sin(2x+y)=3siny,x≠kπ+π/2,x+y≠kπ+π/2(k∈Z)求证:tan(x+y)=2tanx
![sin(2x+y)=3siny,x≠kπ+π/2,x+y≠kπ+π/2(k∈Z)求证:tan(x+y)=2tanx](/uploads/image/z/17535829-13-9.jpg?t=sin%282x%2By%29%3D3siny%2Cx%E2%89%A0k%CF%80%2B%CF%80%2F2%2Cx%2By%E2%89%A0k%CF%80%2B%CF%80%2F2%28k%E2%88%88Z%29%E6%B1%82%E8%AF%81%EF%BC%9Atan%28x%2By%29%3D2tanx)
观察 结论tan(x+y)=2tanx 和条件sin(2x+y)=3siny中
角度的关系,可知
2x+y=(x+y)+(x)
y=(x+y)-(x)
这样可由条件推出结论
条件变为
3sin[(x+y)+x]=sin[(x+y)-x]————根据三角函数两角和公式sin(α+β)=sinαcosβ+cosαsinβ
sin(α-β)=sinαcosβ -cosαsinβ 展开得
3sin(x+y)cosx-3cos(x+y)sinx=sin(x+y)cosx+cos(x+y)sinx,化简
2sin(x+y)cosx=4cos(x+y)sinx
因为x≠kπ+π/2 ,所以cosx≠0
因为x+y≠kπ+π/2,所以cos(x+y)≠0
于是,两边同除以cos(x+y)cosx得
tan(x+y)=2tanx
角度的关系,可知
2x+y=(x+y)+(x)
y=(x+y)-(x)
这样可由条件推出结论
条件变为
3sin[(x+y)+x]=sin[(x+y)-x]————根据三角函数两角和公式sin(α+β)=sinαcosβ+cosαsinβ
sin(α-β)=sinαcosβ -cosαsinβ 展开得
3sin(x+y)cosx-3cos(x+y)sinx=sin(x+y)cosx+cos(x+y)sinx,化简
2sin(x+y)cosx=4cos(x+y)sinx
因为x≠kπ+π/2 ,所以cosx≠0
因为x+y≠kπ+π/2,所以cos(x+y)≠0
于是,两边同除以cos(x+y)cosx得
tan(x+y)=2tanx
tan(x+y)=2tanx(x,x+y≠kπ+π/2,k∈Z),证3siny=sin(2x+y)
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