Find the value of sinA·sin(60°−A)·sin(60°+A)?

Problem: Find the value of \( \sin A \cdot \sin(60^\circ - A) \cdot \sin(60^\circ + A) \)

A) \( \sin 3A \)

B) \( \frac{\sin 3A}{2} \)

C) \( \frac{\sin 3A}{4} \)

D) None of the above

Detailed Solution

We need to simplify \( \sin A \cdot \sin(60^\circ - A) \cdot \sin(60^\circ + A) \).

Step 1: Use the identity \( \sin(P + Q)\sin(P - Q) = \sin^2 P - \sin^2 Q \):

\[ \begin{align*} \sin(60^\circ + A)\sin(60^\circ - A) &= \sin^2 60^\circ - \sin^2 A \\ &= \left(\frac{\sqrt{3}}{2}\right)^2 - \sin^2 A \\ &= \frac{3}{4} - \sin^2 A \end{align*} \]

Step 2: Multiply by \( \sin A \):

\[ \sin A \left(\frac{3}{4} - \sin^2 A\right) = \frac{3}{4}\sin A - \sin^3 A \]

Step 3: Factor out \( \frac{1}{4} \sin A \):

\[ \frac{1}{4} \sin A (3 - 4\sin^2 A) \]

Step 4: Recognize the triple-angle identity \( \sin 3A = 3\sin A - 4\sin^3 A \):

\[ \frac{1}{4} \sin 3A \]

Thus, the value is \( \frac{\sin 3A}{4} \), making option C correct.