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Proof of a Trig Identity

  1. Oct 21, 2006 #1
    This isn't a homework question, it's just for personal enrichment.

    I've been trying to prove that [itex]\tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}[/itex]

    I tried starting off with [tex]\tan\frac{\theta}{2}=\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}[/tex]

    is this even the right way to start the proof? if so could someone point me the right way please? I'd really appreciate it.
    Last edited: Oct 21, 2006
  2. jcsd
  3. Oct 21, 2006 #2
    note that [tex]\tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}[/tex] is equivalent to:

    [tex]\tan\frac{\theta}{2}=\frac{1-\cos \theta}{\sin \theta}[/tex]

    We know that [tex]\frac{1-\cos 2\theta}{\sin 2\theta} = \tan \theta[/tex]

    Can you go from there?
    Last edited: Oct 21, 2006
  4. Oct 21, 2006 #3
    ah I see it now hehe thank you verry much.
  5. Oct 22, 2006 #4
    [tex]\frac{\sin\theta}{1+\cos\theta} = \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}} = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} = \tan\frac{\theta}{2}[/tex]
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