SinA*Sin(60-A)*Sin(60+A)=1/4Sin3A  explain

We are tasked with proving the trigonometric identity:

\[
\sin A \cdot \sin(60^\circ - A) \cdot \sin(60^\circ + A) = \frac{1}{4} \sin(3A)
\]

Let's break this down step by step.

### Step 1: Expand the terms \(\sin(60^\circ - A)\) and \(\sin(60^\circ + A)\)
We can use the **sum and difference formulas** for sine:

\[
\sin(60^\circ - A) = \sin 60^\circ \cos A - \cos 60^\circ \sin A
\]
\[
\sin(60^\circ + A) = \sin 60^\circ \cos A + \cos 60^\circ \sin A
\]

Substituting \(\sin 60^\circ = \frac{\sqrt{3}}{2}\) and \(\cos 60^\circ = \frac{1}{2}\), we get:

\[
\sin(60^\circ - A) = \frac{\sqrt{3}}{2} \cos A - \frac{1}{2} \sin A
\]
\[
\sin(60^\circ + A) = \frac{\sqrt{3}}{2} \cos A + \frac{1}{2} \sin A
\]

### Step 2: Multiply \(\sin(60^\circ - A)\) and \(\sin(60^\circ + A)\)
We now multiply these two expressions:

\[
\sin(60^\circ - A) \cdot \sin(60^\circ + A) = \left( \frac{\sqrt{3}}{2} \cos A - \frac{1}{2} \sin A \right) \cdot \left( \frac{\sqrt{3}}{2} \cos A + \frac{1}{2} \sin A \right)
\]

This is in the form \((x - y)(x + y)\), which is a difference of squares:

\[
= \left(\frac{\sqrt{3}}{2} \cos A\right)^2 - \left(\frac{1}{2} \sin A\right)^2
\]
\[
= \frac{3}{4} \cos^2 A - \frac{1}{4} \sin^2 A
\]

So,
\[
\sin(60^\circ - A) \cdot \sin(60^\circ + A) = \frac{1}{4} (3 \cos^2 A - \sin^2 A)
\]

### Step 3: Substitute this result back into the original expression
The original expression is:

\[
\sin A \cdot \sin(60^\circ - A) \cdot \sin(60^\circ + A)
\]

Substituting the result from Step 2:

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

### Step 4: Use the identity \(\cos^2 A = 1 - \sin^2 A\)
Now, use the Pythagorean identity \(\cos^2 A = 1 - \sin^2 A\) to simplify:

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

### Step 5: Recognize the identity for \(\sin(3A)\)
We know that:

\[
\sin(3A) = 3 \sin A - 4 \sin^3 A
\]

So:

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

### Conclusion
We have successfully shown that:

\[
\sin A \cdot \sin(60^\circ - A) \cdot \sin(60^\circ + A) = \frac{1}{4} \sin(3A)
\]

Thus, the identity is proven.

---

Would you like more details on any part of the proof, or do you have any questions?

### Related Questions:
1. What are some other useful trigonometric identities involving multiple angles?
2. How can the sum and difference formulas be applied to other trigonometric functions?
3. Can this identity be generalized for other angles besides \(60^\circ\)?
4. How does the Pythagorean identity help simplify trigonometric expressions?
5. What is the derivation of the triple angle identity for sine?

### Tip:
When simplifying trigonometric expressions, always look for standard identities like sum/difference, double-angle, or Pythagorean identities that can help reduce complexity.

SinA*Sin(60-A)*Sin(60+A)=1/4Sin3A explain

Explore the step-by-step proof of the trigonometric identity Sin(A) * Sin(60-A) * Sin(60+A) = 1/4 Sin(3A). Learn how to apply sum and difference formulas, the Pythagorean identity, and the triple angle identity for sine. This detailed solution is suitable for high school students in Grades 10-12.

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