Another Great Problem in Trigonometry

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SUMMARY

The problem presented involves the trigonometric identity where if $\cos \alpha + \cos \beta + \cos \gamma = 0$, then the expression $\cos 3 \alpha + \cos 3 \beta + \cos 3 \gamma$ simplifies to $12 \cos \alpha \cos \beta \cos \gamma$. This leads to the conclusion that the value of $\lambda$ is definitively 12. The derivation utilizes the identity for the sum of cubes and the properties of cosine functions in trigonometry.

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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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