In a triangle $ABC$, $\cos 3A + \cos 3B + \cos 3C = 1$, then find any one angle.
HINT: The answer is $\frac{2\pi}{8}$
I have tried these steps and got stuck in the middle.
$$ A +B + C = \pi$$
$$ A + B = \pi – C$$
$$\cos 3A + \cos 3B + \cos 3C = 1 \longrightarrow$$
$$2\cos{\frac{3A + 3B}{2}}\cos{\frac{3A – 3B}{2}} + \cos 3C = 1$$
$$2\cos{\frac{3A + 3B}{2}}\cos{\frac{3A – 3B}{2}} + \cos{(\frac{3\pi}{2}-\frac{3A – 3B}{2})} = 1 \text{ [since $A – B = \pi -C$] } $$
$$2\cos{\frac{3A + 3B}{2}}\cos{\frac{3A – 3B}{2}} – \sin{\frac{3A + 3B}{2}} = 1$$
From here onwards I do not know how to continue. Please check if my method is correct and help me solve the problem. Thanks 🙂
Best Answer
Given $$\cos 3A+\cos3B+\cos3C=1$$ $$\implies\sum\cos3A=1$$ $$\implies\sin\dfrac{3A}{2}\cdot\sin\dfrac{3B}{2}\cdot\sin\dfrac{3C}{2}=0$$
Therefore,$$A=\dfrac{2\pi}{3}\mbox{ or }B=\dfrac{2\pi}{3}\mbox{ or }C=\dfrac{2\pi}{3}$$