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Math Help - equality trigonometry

  1. #1
    arm
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    equality trigonometry

    Let a- b \neq  k\pi :
      a\sin\alpha +b. cos\alpha  = a.sin\beta  +b. cos\beta  =c.

    Prove:  cos^{2}\frac{\alpha -\beta }{2}=\frac{c^{2}}{a^{2}+b^{2}}
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  2. #2
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    \sqrt{a^{2}+b{2}}(\frac {a}{\sqrt {a^{2}+b^{2}}}} \sin \alpha + \frac {b}{\sqrt {a^{2}+b^{2}}}} \cos \alpha) = c

    and

    \sqrt{a^{2}+b{2}}(\frac {a}{\sqrt {a^{2}+b^{2}}}} \sin \beta + \frac {b}{\sqrt {a^{2}+b^{2}}}} \cos \beta) = c

    let's take \theta such that:
    \cos \theta = \frac {b}{\sqrt {a^{2}+b^{2}} }\quad and \quad \sin\theta = \frac {a}{\sqrt {a^{2}+b^{2}}}
      \psi is defined as:
    \cos(\psi)=\frac {c}{\sqrt {a^{2}+b^{2}}}
    We have now:
    \cos (\alpha-\theta)=\cos(\beta-\theta)=\cos(\psi)
    Hence,
    1) \alpha - \theta = \beta-\theta + 2k\pi= \psi+2k'\pi
    2) \alpha - \theta = \beta-\theta + 2k\pi= -(\psi)+2k'\pi
    3)\alpha - \theta = -(\beta-\theta) + 2k\pi= \psi+2k'\pi
    4) \alpha - \theta = -(\beta-\theta) + 2k\pi= -(\psi)+2k'\pi

    1) and 2) are not valid because they violate \alpha - \beta \neq k\pi
    3) and 4) gives us:
    \alpha - \beta = 2 * \psi
    \cos (\frac{\alpha - \beta}{2})=\cos(\psi)
    thus,
    \cos^{2} (\frac{\alpha - \beta}{2})= \frac{c^{2}}{a^2 + b^2}
    Last edited by mestmoha; February 23rd 2011 at 03:45 PM.
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