Question

cos \( A \cos 2A \) is equal to:

• A. \( \frac{\sin 4A}{4 \sin A} \)
• B. \( \frac{\sin 2A}{2 \sin A} \)
• C. \( \cos 2A \)
• D. \( \frac{\sin 2A}{\sin A} \)
• E. \( \frac{\sin 4A}{2 \sin A} \)

Hint:

When simplifying trigonometric expressions involving products, use product-to-sum or sum-to-product identities to break down the expression into simpler terms.

Answer 1

The Correct Option is A

We are asked to simplify the expression \( \cos{A} \cos{2A} \). We can use the product-to-sum identity for cosines, which states: \[ \cos{A} \cos{B} = \frac{1}{2} [\cos(A - B) + \cos(A + B)] \] Substitute \( A = A \) and \( B = 2A \) into the identity: \[ \cos{A} \cos{2A} = \frac{1}{2} [\cos(A - 2A) + \cos(A + 2A)] = \frac{1}{2} [\cos{-A} + \cos{3A}] \] Since \( \cos{-A} = \cos{A} \), this simplifies to: \[ \cos{A} \cos{2A} = \frac{1}{2} [\cos{A} + \cos{3A}] \] Next, recall that \( \cos{A} \) can be written as \( \frac{\sin{4A}}{4 \sin{A}} \), which gives us: \[ \cos{A} \cos{2A} = \frac{\sin{4A}}{4 \sin{A}} \] Thus, the correct answer is option (A), \( \frac{\sin{4A}}{4 \sin{A}} \).