Question

In triangle $ ABC $, given: $$ a = 6,\quad b = 5,\quad c = 4,\quad \text{find } \cos 2A $$

• A. \( -\frac{31}{32} \)
• B. \( -\frac{15}{16} \)
• C. \( \frac{31}{32} \)
• D. \( \frac{15}{16} \)

Hint:

Use cosine law first, then apply double-angle identity \( \cos 2A = 2\cos^2 A - 1 \).

Answer 1

The Correct Option is A

Use cosine law to find \( \cos A \): \[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{5^2 + 4^2 - 6^2}{2 \cdot 5 \cdot 4} = \frac{25 + 16 - 36}{40} = \frac{5}{40} = \frac{1}{8} \] Now use double angle identity: \[ \cos 2A = 2 \cos^2 A - 1 = 2\left( \frac{1}{8} \right)^2 - 1 = 2 \cdot \frac{1}{64} - 1 = \frac{2}{64} - 1 = \frac{1}{32} - 1 = \boxed{ -\frac{31}{32} } \]