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Homework Help: A bit of Trigo

  1. Jan 6, 2006 #1
    By using the Sine formula in triangle ABC, show that:

    [tex]\frac{a+b}{c} = \frac{cos\frac{A-B}{2}}{sin\frac{c}{2}}[/tex].

    I've tried:
    [tex]\frac{2 sin C}{c} = \frac{sin A}{a} + \frac{sin B}{b}[/tex]
    [tex]\frac{2 sin C}{c} = \frac{b sin A + a sin B}{a+b}[/tex]
    [tex]\frac{a+b}{c} = \frac{b sin A + a sin B}{2 sin C}[/tex]
    [tex]\frac{a+b}{c} = \frac{b sin A + a sin B}{4sin\frac{c}{2}cos\frac{c}{2}}[/tex]

    Am I on the right track? Don't really know how to continue.
  2. jcsd
  3. Jan 6, 2006 #2


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    Homework Helper

    No, you are not.
    There's an error when you go from line #1 to line #2. Line #2 should read:
    [tex]\frac{2 sin C}{c} = \frac{b sin A + a sin B}{ab}[/tex] not [tex]\frac{2 sin C}{c} = \frac{b sin A + a sin B}{a+b}[/tex].
    You may want to try this way:
    [tex]\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{a + b}{\sin A + \sin B} = \frac{c}{\sin C}[/tex].
    Can you go from here?
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