In a triangle ABC, a = 4, b = 3, c = 2. What is cos 3C equal to ?

Concept:

For a triangle ABC, with sides a, b, c opposite to angle A, B, C respectively.

Cosine formula-

\(\cos A = {b^2 +c^2 -a^2 \over 2bc}\), \(\cos B = {a^2 +c^2 -b^2 \over 2ac}\) and \(\cos C = {a^2 +b^2 -c^2 \over 2ab}\)

• cos 3x = 4cos³x - 3cos x

Calculation:

Given: a triangle ABC with a = 4, b = 3, c = 2. 

⇒ \(\cos C = {a^2 +b^2 -c^2 \over 2ab} = {4^2 +3^2 -2^2 \over 2(4)(3)}\)

⇒ \(\cos C = {16 + 9 - 4 \over 24} = {21 \over 24}\)

⇒ cos C = \({7 \over 8}\)

So cos 3C = 4cos3C - 3cos C

⇒ \(\cos 3C = 4({7 \over 8})^3 - 3.{7 \over 8}\)

⇒ \(\cos 3C = 4({343 \over 512}) - {21\over 8}\)

⇒ \(\cos 3C = {343 \over 128} - {21\over 8}\)

⇒ \(\cos 3C = {343 - 336\over 128} = {7 \over 128}\)

∴ The correct option is (1).