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Problem with sin(x/2) proof

  1. Oct 13, 2011 #1
    My problem is:
    Proof [itex]sin(\frac{x}{2}) = \pm \sqrt{\frac{1-cos(x)}{2}}[/itex]

    Simple issue really i'd think but i can't come up with a way.

    For starters i'd use however
    [itex]cos^2(x) + sin^2(x) = 1[/itex] identity.

    Which evidently would lead into
    [itex]sin(\frac{x}{2}) = \pm \sqrt{1-cos^2(\frac{x}{2})}[/itex]

    But then i got nothing...
     
  2. jcsd
  3. Oct 13, 2011 #2

    Mentallic

    User Avatar
    Homework Helper

    Do you know the expansion for cos(2x) in terms of sin(x) and cos(x)? From there you would convert this expression solely into terms with sin(x), and finally solve for sin(x).
     
  4. Oct 13, 2011 #3
    Ahhh i get it!
    [itex]
    cos(2\frac{x}{2}) = cos^2(\frac{x}{2})-sin^2(\frac{x}{2})
    [/itex]
    [itex]
    cos^2(\frac{x}{2}) = cos(2\frac{x}{2})-sin^2(\frac{x}{2})
    [/itex]

    Thus
    [itex]
    sin^2(\frac{x}{2}) = 1-cos^2(\frac{x}{2})
    [/itex]
    [itex]
    sin^2(\frac{x}{2}) = 1-cos(x)-sin^2(\frac{x}{2})
    [/itex]

    And so
    [itex]
    sin(\frac{x}{2}) = \pm \sqrt{\frac{1-cos(x)}{2}}
    [/itex]
    Thanks!
     
  5. Oct 13, 2011 #4

    Mark44

    Staff: Mentor

    I'm not sure you do.
    No,
    [itex]cos^2(\frac{x}{2}) = cos(2\frac{x}{2}) + sin^2(\frac{x}{2})[/itex]
     
  6. Oct 13, 2011 #5
    That was a typo but anyway... :)
     
  7. Oct 13, 2011 #6

    Mentallic

    User Avatar
    Homework Helper

    Nice work :smile:
     
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