锐角三角形ABC,证:tanAtanB相乘大于1
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锐角三角形ABC,证:tanAtanB相乘大于1
锐角三角形ABC,证:tanAtanB相乘大于1
锐角三角形ABC,证:tanAtanB相乘大于1
因为△ABC为锐角三角形
故C<90°,tanA>0,tanB>0,180>A+B=180-C>90
故tanA+tanB>0,tan(A+B)=(tanA+tanB)/(1-tanAtanB)<0
即得tanAtanB>1
tanAtanB=sinA*sinB/cosA*cosB
积化和差
tanAtanB=sinA*sinB/cosA*cosB=-[cos(A+B)-cos(A-B)] / [cos(A+B)+cos(A-B)]
=-[cos(π-C)-cos(A-B)] / [cos(π-C)+cos(A-B)]
=[cosC+cos(A-B)] / [cos(A-B)-cosC]...
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tanAtanB=sinA*sinB/cosA*cosB
积化和差
tanAtanB=sinA*sinB/cosA*cosB=-[cos(A+B)-cos(A-B)] / [cos(A+B)+cos(A-B)]
=-[cos(π-C)-cos(A-B)] / [cos(π-C)+cos(A-B)]
=[cosC+cos(A-B)] / [cos(A-B)-cosC]
锐角三角形ABC中∠C<90° -90°<∠(A-B) <90° cosC>0 cos(A-B)>0
又∠(A-B) <∠C cos(A-B)>cosC 故恒有cosC+cos(A-B)>cos(A-B)-cosC 则tanAtanB=[cosC+cos(A-B)] / [cos(A-B)-cosC]>1
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