In a triangle ABC, a=4, b=3, c=2. What is \cos 3C equal to ?
- A. \frac{7}{128} ✓
- B. \frac{11}{128}
- C. \frac{7}{64}
- D. \frac{11}{64}
Correct Answer: A. \frac{7}{128}
Explanation
By the cosine rule, \cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{16+9-4}{2(4)(3)} = \frac{21}{24} = \frac{7}{8}. The identity for \cos 3C is 4\cos^3 C - 3\cos C. Substituting gives 4(\frac{7}{8})^3 - 3(\frac{7}{8}) = 4(\frac{343}{512}) - \frac{21}{8} = \frac{343}{128} - \frac{336}{128} = \frac{7}{128}.