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Thread: Super Challenging Trigo question

  1. #1
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    Super Challenging Trigo question

    Suppose $\displaystyle 0\leq\alpha<\beta<\gamma<2\pi$ and
    $\displaystyle cos\alpha+cos\beta+cos\gamma=sin\alpha+sin\beta+si n\gamma=0$, find the value of $\displaystyle (\beta-\alpha)$
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  2. #2
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    Hello acc100jt
    Quote Originally Posted by acc100jt View Post
    Suppose $\displaystyle 0\leq\alpha<\beta<\gamma<2\pi$ and
    $\displaystyle cos\alpha+cos\beta+cos\gamma=sin\alpha+sin\beta+si n\gamma=0$, find the value of $\displaystyle (\beta-\alpha)$
    Re-write:

    $\displaystyle \left\{\begin{array}{l}\cos\alpha +\cos\beta = -\cos\gamma\\ \sin\alpha+\sin\beta = -\sin\gamma\end{array} \right. \$

    Square and add:

    $\displaystyle (\cos\alpha +\cos\beta)^2 + (\sin\alpha +\sin\beta)^2 = \cos^2\gamma + \sin^2\gamma =1$

    $\displaystyle \Rightarrow \cos^2\alpha +2\cos\alpha\cos\beta + \cos^2\beta +\sin^2\alpha +2\sin\alpha\sin\beta + \sin^2\beta=1$

    $\displaystyle \Rightarrow 2 + 2\cos(\beta - \alpha) = 1$

    $\displaystyle \Rightarrow \cos(\beta - \alpha) = -\tfrac12$

    $\displaystyle \Rightarrow \beta - \alpha = 2n\pi \pm \frac{2\pi}{3}$

    Grandad
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