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Math Help - Equality !!

  1. #1
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    Equality !!

    Hi !!

    \alpha , \beta and \gamma the angles of a triangle , prove that :




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  2. #2
    MHF Contributor red_dog's Avatar
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    \cos\alpha+\cos\beta+1-\cos\gamma=2\cos\frac{\alpha+\beta}{2}\cos\frac{\a  lpha-\beta}{2}+2\sin^2\frac{\gamma}{2}=

    =2\sin\frac{\gamma}{2}\cos\frac{\alpha-\beta}{2}+2\sin^2\frac{\gamma}{2}=2\sin\frac{\gamm  a}{2}\left(\cos\frac{\alpha-\beta}{2}+\sin\frac{\gamma}{2}\right)=

    =2\sin\frac{\gamma}{2}\left(\cos\frac{\alpha-\beta}{2}+\cos\frac{\alpha+\beta}{2}\right)=4\sin\  frac{\gamma}{2}\cos\frac{\alpha}{2}\cos\frac{\beta  }{2}

    Then \frac{\cos\alpha+\cos\beta}{\cos\alpha+\cos\beta+1-\cos\gamma}=\frac{2\sin\frac{\gamma}{2}\cos\frac{\  alpha-\beta}{2}}{4\sin\frac{\gamma}{2}\cos\frac{\alpha}{  2}\cos\frac{\beta}{2}}=\frac{\cos\frac{\alpha-\beta}{2}}{2\cos\frac{\alpha}{2}\cos\frac{\beta}{2  }}

    \frac{\cos\gamma+\cos\beta}{\cos\gamma+\cos\beta+1-\cos\alpha}=\frac{2\sin\frac{\alpha}{2}\cos\frac{\  gamma-\beta}{2}}{4\sin\frac{\alpha}{2}\cos\frac{\gamma}{  2}\cos\frac{\beta}{2}}=\frac{\cos\frac{\gamma-\beta}{2}}{2\cos\frac{\gamma}{2}\cos\frac{\beta}{2  }}

    \frac{\cos\alpha+\cos\gamma}{\cos\alpha+\cos\gamma  +1-\cos\beta}=\frac{2\sin\frac{\beta}{2}\cos\frac{\al  pha-\gamma}{2}}{4\sin\frac{\beta}{2}\cos\frac{\alpha}{  2}\cos\frac{\gamma}{2}}=\frac{\cos\frac{\alpha-\gamma}{2}}{2\cos\frac{\alpha}{2}\cos\frac{\gamma}  {2}}

    Then the left side member of the equality is

    \frac{\cos\frac{\alpha-\beta}{2}}{2\cos\frac{\alpha}{2}\cos\frac{\beta}{2  }}+\frac{\cos\frac{\beta-\gamma}{2}}{2\cos\frac{\beta}{2}\cos\frac{\gamma}{  2}}+\frac{\cos\frac{\gamma-\alpha}{2}}{2\cos\frac{\gamma}{2}\cos\frac{\alpha}  {2}}=

    =\frac{\cos\frac{\gamma}{2}\cos\frac{\alpha-\beta}{2}+\cos\frac{\alpha}{2}\cos\frac{\beta-\gamma}{2}+\cos\frac{\beta}{2}\cos\frac{\gamma-\alpha}{2}}{2\cos\frac{\alpha}{2}\cos\frac{\beta}{  2}\cos\frac{\gamma}{2}}=

    =\frac{\cos\frac{\gamma+\alpha-\beta}{2}+\cos\frac{\gamma-\alpha+\beta}{2}+\cos\frac{\alpha+\beta-\gamma}{2}+\cos\frac{\alpha-\beta+\gamma}{2}+\cos\frac{\beta+\gamma-\alpha}{2}+\cos\frac{\beta-\gamma+\alpha}{2}}{4\cos\frac{\alpha}{2}\cos\frac{  \beta}{2}\cos\frac{\gamma}{2}}=

    =\frac{2\cos\left(\frac{\pi}{2}-\beta\right)+2\cos\left(\frac{\pi}{2}-\alpha\right)+2\cos\left(\frac{\pi}{2}-\gamma\right)}{4\cos\frac{\alpha}{2}\cos\frac{\bet  a}{2}\cos\frac{\gamma}{2}}=

    =\frac{\sin\alpha+\sin\beta+\sin\gamma}{2\cos\frac  {\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\gamma}{2}  }=\frac{4\cos\frac{\alpha}{2}\cos\frac{\beta}{2}\c  os\frac{\gamma}{2}}{2\cos\frac{\alpha}{2}\cos\frac  {\beta}{2}\cos\frac{\gamma}{2}}=2
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