Question

In a triangle ABC , cos 3A+ cos 3B+ cos 3C = 1 ,then find any one angle.

Answer 1

Cos(3A) + Cos(3B) - Cos(3A+3B) = 1

which we can expand into:

Cos(3A) + Cos(3B) - Cos(3A)Cos(3B) + Sin(3A)Sin(3B) = 1

1 - (1-Cos(3A))(1-Cos(3B)) + Sin(3A)Sin(3B) = 1

that is, we have:

(1-Cos(3A))(1-Cos(3B) = Sin(3A)Sin(3B)

Tan{(3/2)A) = Cot((3/2)B)

Thus, we know that (3/2)(A+B) = 90°, or A+B = 60°, and therefore C = 120°