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: Combine two factors
Use \(\cos X\cos Y = \frac12[\cos(X+Y)+\cos(X-Y)]\) on \(\cos(60^{\circ}-A)\cos(60^{\circ}+A)\).

Step 2: Simplify
\[ \cos(60^{\circ}-A)\cos(60^{\circ}+A) = \frac12[\cos120^{\circ} + \cos 2A] = \frac12\left[-\frac12 + 2\cos^2A - 1 \right] \]
\[ = \cos^2A - \frac34 \]

Step 3: Multiply by cos A
\[ \cos A\left(\cos^2A - \frac34\right) = \frac14(4\cos^3A - 3\cos A) = \frac14\cos3A \]

Step 4: Check
Take \(A=0\): LHS \(=\cos60^{\circ}\cdot1\cdot\cos60^{\circ} = \frac14\). Option (A) gives \(\frac14\cos 0 = \frac14\). Option (C) gives \(1\), which fails.

Final Answer:
The value is one quarter of cos 3A. \[ \boxed{\text{(A)}\ \frac14\cos3A} \]