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Math Help - Measure-angles

  1. #1
    Super Member dhiab's Avatar
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    Measure-angles

    a,b,c measure angles of triangle.
    Prove :
    \tan \left( {\frac{a}{2}} \right) + \tan \left( {\frac{b}{2}} \right) + \tan \left( {\frac{c}{2}} \right) = 4\frac{{1 + \sin \left( {\frac{a}{2}} \right)\sin \left( {\frac{b}{2}} \right)\sin \left( {\frac{c}{2}} \right)}}{{\sin a + \sin b + \sin c}}<br />
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  2. #2
    MHF Contributor red_dog's Avatar
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    \tan\frac{a}{2}+\tan\frac{a}{2}+\tan\frac{c}{2}=\f  rac{\sin\frac{a}{2}}{\cos\frac{a}{2}}+\frac{\sin\f  rac{b}{2}}{\cos\frac{b}{2}}+\frac{\sin\frac{c}{2}}  {\cos\frac{c}{2}}=

    =\frac{\sin\frac{a+b}{2}}{\cos\frac{a}{2}\cos\frac  {b}{2}}+\frac{\sin\frac{c}{2}}{\cos\frac{c}{2}}=

    =\frac{\cos\frac{c}{2}}{\cos\frac{a}{2}\cos\frac{b  }{2}}+\frac{\sin\frac{c}{2}}{\cos\frac{c}{2}}=

    =\frac{\cos^2\frac{c}{2}+\sin\frac{c}{2}\cos\frac{  a}{2}\cos\frac{b}{2}}{\cos\frac{a}{2}\cos\frac{b}{  2}\cos\frac{c}{2}}=

    =\frac{1-\sin^2\frac{c}{2}+\sin\frac{c}{2}\cos\frac{a}{2}\c  os\frac{b}{2}}{\cos\frac{a}{2}\cos\frac{b}{2}\cos\  frac{c}{2}}=

    =\frac{1+\sin\frac{c}{2}\left(\cos\frac{a}{2}\cos\  frac{b}{2}-\cos\frac{a+b}{2}\right)}{\cos\frac{a}{2}\cos\frac  {b}{2}\cos\frac{c}{2}}=

    =\frac{1+\sin\frac{a}{2}\sin\frac{b}{2}\sin\frac{c  }{2}}{\cos\frac{a}{2}\cos\frac{b}{2}\cos\frac{c}{2  }}=

    But it is well known that \sin a+\sin b+\sin c=4\cos\frac{a}{2}\cos\frac{b}{2}\cos\frac{c}{2}=
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  3. #3
    Senior Member pacman's Avatar
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    reddog, it seems too easy for you . . . . well done
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