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Thread: equality trigonometry

  1. #1
    arm
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    equality trigonometry

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

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

    and

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

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

    1) and 2) are not valid because they violate $\displaystyle \alpha - \beta \neq k\pi$
    3) and 4) gives us:
    $\displaystyle \alpha - \beta = 2 * \psi$
    $\displaystyle \cos (\frac{\alpha - \beta}{2})=\cos(\psi)$
    thus,
    $\displaystyle \cos^{2} (\frac{\alpha - \beta}{2})= \frac{c^{2}}{a^2 + b^2}$
    Last edited by mestmoha; Feb 23rd 2011 at 02:45 PM.
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