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Half angle formula,

  1. Dec 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Can someone please explain how my book got [itex] sin^{2} \theta = \frac{1-cos2\theta}{2} [/itex] and [itex] cos^{2}\theta = \frac{1+cos2\theta}{2} [/itex]

    As I thought the half angle formula's were [itex] sin \frac{\theta}{2} = \sqrt{\frac{1-cos\theta}{2}} [/itex]

    [itex] cos\frac{\theta}{2} = \sqrt\frac{1+cos\theta}{2}} [/itex]

    So how can it also be the top one's aswell?

  2. jcsd
  3. Dec 20, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Replace θ with 2θ.
  4. Dec 20, 2009 #3

    Oh I see, can't believe I missed that.

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