1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: A bit of Trigo
  1. Trigo doubt (Replies: 4)

  2. Trigo/Geom. problem (Replies: 7)