If sin A sin`(60^@-A)sin(60^@+A)=k` sin 3A , then what is k equal to?

Options

A. `1/4`
B. `1/2`
C. 1
D. 4

Answer

To solve the equation \( \sin A \sin(60^\circ - A) \sin(60^\circ + A) = k \sin 3A \), we will follow these steps:

### Step 1: Use the Product-to-Sum Formulas
We can use the product-to-sum identities to simplify \( \sin(60^\circ - A) \sin(60^\circ + A) \).

\[ \sin(60^\circ - A) \sin(60^\circ + A) = \frac{1}{2} \left[ \cos(-A) - \cos(120^\circ) \right] \]

Since \( \cos(-A) = \cos A \) and \( \cos(120^\circ) = -\frac{1}{2} \), we have:

\[ \sin(60^\circ - A) \sin(60^\circ + A) = \frac{1}{2} \left[ \cos A + \frac{1}{2} \right] = \frac{1}{2} \cos A + \frac{1}{4} \]

### Step 2: Substitute Back into the Original Equation
Now we substitute this back into the original equation:

\[ \sin A \left( \frac{1}{2} \cos A + \frac{1}{4} \right) = k \sin 3A \]

### Step 3: Expand the Left Side
Expanding the left side gives us:

\[ \frac{1}{2} \sin A \cos A + \frac{1}{4} \sin A = k \sin 3A \]

### Step 4: Use the Identity for \( \sin 3A \)
Recall that \( \sin 3A = 3 \sin A - 4 \sin^3 A \). Thus, we can write:

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

### Step 5: Factor Out \( \sin A \)
Factoring out \( \sin A \) from both sides gives:

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

### Step 6: Set Up for \( k \)
To find \( k \), we can evaluate this equation at a specific angle, for example, \( A = 0 \):

\[ k(3 - 0) = \frac{1}{2} \cdot 1 + \frac{1}{4} \implies 3k = \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4} \]

### Step 7: Solve for \( k \)
Now we can solve for \( k \):

\[ k = \frac{3/4}{3} = \frac{1}{4} \]

Thus, the value of \( k \) is:

\[ \boxed{\frac{1}{4}} \]