# Super Challenging Trigo question

• Apr 21st 2009, 06:51 PM
acc100jt
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 $\displaystyle (\beta-\alpha)$
• Apr 21st 2009, 10:47 PM
Hello acc100jt
Quote:

Originally Posted by acc100jt
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. \$

$\displaystyle (\cos\alpha +\cos\beta)^2 + (\sin\alpha +\sin\beta)^2 = \cos^2\gamma + \sin^2\gamma =1$
$\displaystyle \Rightarrow \cos^2\alpha +2\cos\alpha\cos\beta + \cos^2\beta +\sin^2\alpha +2\sin\alpha\sin\beta + \sin^2\beta=1$
$\displaystyle \Rightarrow 2 + 2\cos(\beta - \alpha) = 1$
$\displaystyle \Rightarrow \cos(\beta - \alpha) = -\tfrac12$
$\displaystyle \Rightarrow \beta - \alpha = 2n\pi \pm \frac{2\pi}{3}$