请教一个问题1求证(tanαtan2α)/(tan2α-tanα)+√3(sin²α-cos²α)=
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请教一个问题1求证(tanαtan2α)/(tan2α-tanα)+√3(sin²α-cos²α)=2(2α-π/3)
![请教一个问题1求证(tanαtan2α)/(tan2α-tanα)+√3(sin²α-cos²α)=](/uploads/image/z/19008330-42-0.jpg?t=%E8%AF%B7%E6%95%99%E4%B8%80%E4%B8%AA%E9%97%AE%E9%A2%981%E6%B1%82%E8%AF%81%28tan%CE%B1tan2%CE%B1%29%2F%28tan2%CE%B1-tan%CE%B1%29%EF%BC%8B%E2%88%9A3%EF%BC%88sin%26sup2%3B%CE%B1-cos%26sup2%3B%CE%B1%EF%BC%89%3D)
左边=(sina/cosa*sin2a/cos2a)/(sin2a/cos2a-sina/cosa)-√3(cos²a-sin²a)
上下乘cosacos2a
=(sinasin2a/(sin2acosa-sinacos2a)-√3(cos²a-sin²a)
=(sinasin2a)/sin(2a-a)-√3cos2a
=sin2a-√3cos2a
=2(sin2a*1/2-cos2a√3/2)
=2(sin2acosπ/3-cos2asinπ/3)
=2sin(2a-π/3)
=右边
命题得证
上下乘cosacos2a
=(sinasin2a/(sin2acosa-sinacos2a)-√3(cos²a-sin²a)
=(sinasin2a)/sin(2a-a)-√3cos2a
=sin2a-√3cos2a
=2(sin2a*1/2-cos2a√3/2)
=2(sin2acosπ/3-cos2asinπ/3)
=2sin(2a-π/3)
=右边
命题得证
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