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Math Help - Tangets of half angles in a triangle

  1. #1
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    Tangets of half angles in a triangle

    Prove that in a triangle with angles \alpha, \beta and \gamma:
    \tan\frac{\alpha}2\tan\frac{\beta}2+\tan\frac{\bet  a}2\tan\frac{\gamma}2+\tan\frac{\gamma}2\tan\frac{  \alpha}2=1.
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  2. #2
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    Quote Originally Posted by james_bond View Post
    Prove that in a triangle with angles \alpha, \beta and \gamma:
    \tan\frac{\alpha}2\tan\frac{\beta}2+\tan\frac{\bet  a}2\tan\frac{\gamma}2+\tan\frac{\gamma}2\tan\frac{  \alpha}2=1.
    I will use a,b,c instead because it is easier to type in LaTeX.

    If a,b,c are angles of a triangle then a+b+c = \pi.

    That means, \tan \frac c2 = \tan \left( \frac{\pi - a - b}2 \right) = \tan \left( \frac{\pi}2 - \frac{a+b}2 \right) = \cot \frac{a+b}2

    Thus, \sum_{cyc} \tan \frac a2 \tan \frac b2 = \tan \frac a2 \tan \frac b2 + \tan \frac a2 \cot \frac{a+b}2 + \tan \frac b2 \cot \frac{a+b}2

    Factor, \tan \frac a2  \tan \frac b2 + \cot \frac{a+b}2 \left( \tan \frac a2 + \tan \frac b2 \right)

    Use identity, \tan \frac a2 \tan \frac b2 + \left( \frac{1 - \tan \frac a2 \tan \frac b2}{\tan \frac a2 + \tan \frac b2} \right) \cdot  \left( \tan \frac a2 + \tan \frac b2 \right) = \tan \frac a2 \tan \frac b2 + 1 - \tan \frac a2  \tan \frac b2 = 1
    Last edited by ThePerfectHacker; January 13th 2008 at 01:36 PM.
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  3. #3
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    Thank you very much, just a little typo at the end...
    \dots= \tan \frac a2 \tan \frac b2 + 1 - \tan \frac a2 \tan \frac b2 = 1
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  4. #4
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    Quote Originally Posted by james_bond View Post
    Thank you very much, just a little typo at the end...
    \dots= \tan \frac a2 \tan \frac b2 + 1 - \tan \frac a2 \tan \frac b2 = 1
    Fixed.
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  5. #5
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    Sorry but in a triange a+b+c=\pi. Am I wrong? .
    (Of course the rest seems to be correct...)
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  6. #6
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    Quote Originally Posted by james_bond View Post
    Sorry but in a triange a+b+c=\pi. Am I wrong? .
    (Of course the rest seems to be correct...)
    Yes, another mistake. I meant they add up to \pi.
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