Welcome to our community

Be a part of something great, join today!

Trigonometry Another Great Problem in Trigonometry

DrunkenOldFool

New member
Feb 6, 2012
20
If $\cos \alpha +\cos \beta + \cos \gamma=0$ and $\cos 3 \alpha +\cos 3\beta +\cos 3\gamma = \lambda \cos \alpha \cos \beta \cos \gamma$. What is the value of $\lambda$?
 

sbhatnagar

Active member
Jan 27, 2012
95
If $\cos \alpha +\cos \beta + \cos \gamma=0$ and $\cos 3 \alpha +\cos 3\beta +\cos 3\gamma = \lambda \cos \alpha \cos \beta \cos \gamma$. What is the value of $\lambda$?
\[\begin{align*}\cos 3 \alpha +\cos 3\beta +\cos 3\gamma &= (4\cos^3 \alpha -3\cos \alpha)+(4\cos^3 \beta -3\cos \beta)+(4\cos^3 \gamma -3\cos \gamma) \\ &= 4(\cos^3 \alpha +\cos^3 \beta +\cos^3 \gamma)-3(\underbrace{\cos \alpha +\cos \beta + \cos \gamma}_{=0}) \\ &= 4(\cos^3 \alpha +\cos^3 \beta +\cos^3 \gamma)\end{align*}\]

Note that when $a+b+c=0$, $a^3+b^3+c^3=3abc$.

\[\begin{align*}\cos 3 \alpha +\cos 3\beta +\cos 3\gamma &= 4(3\cos \alpha \cos \beta \cos \gamma) \\ &= 12\cos \alpha \cos \beta \cos \gamma\end{align*}\]

Therefore $\lambda =12$.