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!

Trigonometry I

  1. Jun 22, 2008 #1
    1. The problem statement, all variables and given/known data

    Prove
    [tex]\frac{sin 2 \alpha}{1+cos 2 \alpha}-\frac{cos \alpha}{1+cos \alpha}=tan \frac{ \alpha}{2}[/tex]




    Use half and double angle formulas, I get;

    The given expression is equal to:
    [tex]\frac{(2 tan \frac{\alpha}{2})(1+tan^{2} \frac{\alpha}{2})}{1-tan^{2}\frac{\alpha}{2}}+2 tan \frac{\alpha}{2}+tan^{2}\frac{\alpha}{2}-1[/tex]

    Please help me simplify this to only [tex]tan \frac{\alpha}{2}[/tex]
     
    Last edited: Jun 22, 2008
  2. jcsd
  3. Jun 22, 2008 #2
    That form looks quite complicated.

    Why don't you try expanding the RHS instead of the LHS using the half angle formula
    [tex]tan \frac{\alpha}{2}= \pm \sqrt_{\frac{1-cos \alpha}{1+cos \alpha} [/tex] [tex]= \frac{sin\alpha}{1+cos\alpha}[/tex]
     
    Last edited: Jun 22, 2008
  4. Jun 22, 2008 #3
    This implies that:
    [tex]2 tan \alpha - \frac{cos \alpha}{1+ cos \alpha}=\frac{sin \alpha}{1+ cos \alpha}[/tex] Isnt it?

    But now the denominators are different. How will I proceed?

    Thanks
     
  5. Jun 22, 2008 #4
    Oh darn. I didn't turn out like I hoped it would. Lol. Let's try your original method.

    Let's see, by the double-angle formula,
    [tex]tan(2\alpha)=\frac{2tan \alpha}{1-tan^2 \alpha} [/tex] (*)

    therefore, your original LHS
    then for the [tex]\frac{(2 tan \frac{\alpha}{2})(1+tan^{2} \frac{\alpha}{2})}{1-tan^{2}\frac{\alpha}{2}}[/tex] term


    By (*) [tex]1-tan^2 \frac{\alpha}{2}= \frac{2tan \frac{\alpha}{2}}{tan \alpha}[/tex], so the [tex]\frac{(2 tan \frac{\alpha}{2})[/tex] should cancel

    Edit:
    This simplifies the LHS into
    [tex]tan \alpha \left( 1+tan^2 \frac{\alpha}{2} \right) + +2 tan \frac{\alpha}{2}+tan^{2}\frac{\alpha}{2}-1[/tex]
     
    Last edited: Jun 22, 2008
  6. Jun 22, 2008 #5

    Defennder

    User Avatar
    Homework Helper

    Did you transcribe the question correctly? It appears that the "identity" doesn't hold for [tex]\alpha = \frac{\pi}{4}[/tex]
     
  7. Jun 23, 2008 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi ritwik06! :smile:
    ooh! :cry:

    Golden rule: whnever you see (1 + cos) or (1 - cos), use the standard trigonometric identities for them! :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?