已知ABC是三角形的内角,求证tanA/2*tanB/2+tanB/2*tanC/2+tanC/2*tanA/2=1
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![已知ABC是三角形的内角,求证tanA/2*tanB/2+tanB/2*tanC/2+tanC/2*tanA/2=1](/uploads/image/z/987436-28-6.jpg?t=%E5%B7%B2%E7%9F%A5ABC%E6%98%AF%E4%B8%89%E8%A7%92%E5%BD%A2%E7%9A%84%E5%86%85%E8%A7%92%2C%E6%B1%82%E8%AF%81tanA%2F2%2AtanB%2F2%2BtanB%2F2%2AtanC%2F2%2BtanC%2F2%2AtanA%2F2%3D1)
已知ABC是三角形的内角,求证tanA/2*tanB/2+tanB/2*tanC/2+tanC/2*tanA/2=1
已知ABC是三角形的内角,求证tanA/2*tanB/2+tanB/2*tanC/2+tanC/2*tanA/2=1
已知ABC是三角形的内角,求证tanA/2*tanB/2+tanB/2*tanC/2+tanC/2*tanA/2=1
A/2+B/2+C/2=90°
A/2=90°-(B/2+C/2)
tanA/2 = tan(90°-(B/2+C/2))
= cot(B/2+C/2)=1/tan(B/2+C/2)=(1-tanB/2tanC/2)/(tanB/2+tanC/2)
tanA/2 (tanB/2+tanC/2) = 1-tanB/2tanC/2
tanA/2 tanB/2 + tanA/2tanC/2 = 1-tanB/2tanC/2
tanA/2 tanB/2 + tanB/2tanC/2 + tanA/2tanC/2 = 1
将tanB/2提出来,tanA/2+tanC/2=tan(A/2+C/2)*(1-tanA/2*tanC/2)
这一题是错题,没法证明。
随便找个例子就可以推翻:
假设A=B=C=60'
因为tan60'=根3
所以原式等于9/4
不等于1
所以,不成立