# Thread: Super Challenging Trigo question

1. ## 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 $\displaystyle (\beta-\alpha)$

2. Hello acc100jt
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. \$

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

### trigo challenging question

Click on a term to search for related topics.