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)
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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