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Trig trig trig

  1. Sep 29, 2005 #1
    hey everyone,
    i have to show that in a triangle, there is : (A, B, C are the angles of that triangle)
    [tex] \sin \left( 1/2\,B \right) \cos \left( 1/2\,C \right) +\sin \left( 1/2

    \,C \right) \cos \left( 1/2\,B \right) =\cos \left( 1/2\,A \right) [/tex]

    for this one, here is what i got to...

    [tex] 1/4\,\sqrt {2-2\,{\it cosB}}\sqrt {2+2\,{\it cosC}}+1/4\,\sqrt {2-2\,{

    \it cosC}}\sqrt {2+2\,{\it cosB}} [/tex]


    my question is, i dont have any A in this equation above, and i have to prove that it is equal to cos(a/2)!! i know that cos(a)=cos(pi-(b+c) ... please if someone can help me with that!

    for the second one, we have : [tex] \left( \cos \right) \,\alpha={\frac {a}{b+c}} [/tex]

    and [tex] \left( \cos \right) \,\beta={\frac {b}{c+a}}
    and \left( \cos \right) \,\gamma={\frac {c}{a+b}} [/tex]

    and we have to show that:
    [tex] 1/2\,{\tan}^{2}\alpha+1/2\,{\tan}^{2}\beta+1/2\,{\tan}^{2}\gamma = 1 [/tex]

    this one, i erally dont know what to do! if someone can help me out! or maybe give me hints...

    i appreciate your help!!
     
  2. jcsd
  3. Sep 29, 2005 #2

    TD

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    You can use the fact that

    [tex]\tan ^2 \theta = \frac{{\sin ^2 \theta }}
    {{\cos ^2 \theta }} = \frac{{1 - \cos ^2 \theta }}
    {{\cos ^2 \theta }} = \frac{1}
    {{\cos ^2 \theta }} - 1[/tex]
     
  4. Sep 29, 2005 #3
    Sorry... for the second one, there is a mistake (I am still learning how to us LaTex so, yeah) here is the correct form: I am not gonna use LaTex, I can't get what I want...

    tan²(α/2) + tan²(β/2) + tan²(γ/2) = 1

    Thanks,
     
  5. Sep 29, 2005 #4

    TD

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    Then you can use the fact that:

    [tex]\tan \left( {\frac{\theta }
    {2}} \right) = \frac{{\sin \theta }}
    {{1 - \cos \theta }} \Leftrightarrow \tan ^2 \left( {\frac{\theta }
    {2}} \right) = \frac{{\sin ^2 \theta }}
    {{\left( {1 - \cos \theta } \right)^2 }} = \frac{{1 - \cos ^2 \theta }}
    {{\left( {1 - \cos \theta } \right)^2 }} = \frac{{\left( {1 - \cos \theta } \right)\left( {1 + \cos \theta } \right)}}
    {{\left( {1 - \cos \theta } \right)^2 }} = \frac{{1 + \cos \theta }}
    {{1 - \cos \theta }}[/tex]
     
  6. Sep 29, 2005 #5
    okay, so iam here:
    [tex]
    \left( {\frac {a}{b+c}}+1 \right) \left( -{\frac {a}{b+c}}+1

    \right) ^{-1}+ \left( 1+{\frac {b}{a+c}} \right) \left( 1-{\frac {b}

    {a+c}} \right) ^{-1}+ \left( 1+{\frac {c}{a+b}} \right) \left( 1-{

    \frac {c}{a+b}} \right) ^{-1}

    [/tex]

    is there any way that this can be equal to 0 !!

    PS: sorry about messing up the equation in latex, but i am sure u still can get what i did (i am using maple 10 to convert to latex, i think there is a problem there)
     
  7. Sep 29, 2005 #6

    TD

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    I calculated it too and I don't think it simplifies to 1, although I can't find an error in my derivation for the tangents to cosines.
     
  8. Sep 29, 2005 #7
    Okay, and for the first one, I noticed that what I had is a sine rule... it's equal tu sin((b+c)/2) but, where do i get it to get equal to cos (a/2) ... :huh:
     
  9. Sep 29, 2005 #8

    TD

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    Very good, but we know that [itex]a+b+c=180[/itex] so:

    [tex]\sin \left( {\frac{{a + b}}
    {2}} \right) = \sin \left( {\frac{{180 - c}}
    {2}} \right) = \sin \left( {90 - \frac{c}
    {2}} \right) = \cos \left( { - \frac{c}
    {2}} \right) = \cos \left( {\frac{c}
    {2}} \right)[/tex]
     
  10. Sep 29, 2005 #9
    yup, thanks!
     
  11. Sep 29, 2005 #10

    TD

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    No problem :smile:

    Never forget to use everythings that's given.
     
  12. Sep 29, 2005 #11
    but anyone, can help me with the second problem?! pliiiiz
     
  13. Sep 30, 2005 #12
    ok i got it. never mind everyone!
     
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