1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Trigonometry I

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

    [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?

  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

    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


    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


    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:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook