Question

The value of \(cos(60^{\circ}-A)\cdot cosA\cdot cos(60^{\circ}+A)\) is

• A. \(\frac{1}{4}cos3A\)
• B. \(sin3A\)
• C. \(cos3A\)
• D. \(\frac{1}{4}sin3A\)

Hint:

Combine the first and last factors using the product-to-sum formula.

Answer 1

The Correct Option is A

Step 1: Use the identity for cos(60+A) cos(60-A)
$\cos(60^{\circ}+A)\cos(60^{\circ}-A) = \cos^2 60^{\circ} - \sin^2 A = \frac14 - \sin^2A$.

Step 2: Multiply by cos A
$\cos A(\tfrac14 - \sin^2A) = \cos A\cdot\frac{1 - 4\sin^2A}{4} = \frac{\cos A(4\cos^2A - 3)}{4}$.

Step 3: Triple angle
$4\cos^3A - 3\cos A = \cos3A$, so the result is $\frac14\cos3A$. Option (A).

Final Answer:
One quarter cos 3A. \[ \boxed{\text{(A)}\ \frac14\cos3A} \]