MHB Another Great Problem in Trigonometry

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The problem involves finding the value of λ given the equations involving cosines of angles α, β, and γ. It is established that if cos α + cos β + cos γ = 0, then cos 3α + cos 3β + cos 3γ can be expressed as 4 times the sum of the cubes of the cosines. Using the identity for the sum of cubes when the angles sum to zero, it simplifies to 12 times the product of the cosines. Consequently, the value of λ is determined to be 12. This conclusion highlights a specific relationship in trigonometric identities.
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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 $\lambda$?
 
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DrunkenOldFool said:
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 $\lambda =12$.
 
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