If `sin theta=(1)/(2) (a+(1)/(a)) and sin 3 theta=(k)/(2)(a^(3)+(1)/(a^(3)))`, then `k+6` is equal to :

Options

A. 3
B. 4
C. 5
D. -4

Answer

To solve the problem, we start with the given equations:

1. \( \sin \theta = \frac{1}{2} \left( a + \frac{1}{a} \right) \)
2. \( \sin 3\theta = \frac{k}{2} \left( a^3 + \frac{1}{a^3} \right) \)

We need to find the value of \( k + 6 \).

### Step 1: Use the formula for \( \sin 3\theta \)

We know that:
\[ \sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta \]

### Step 2: Substitute \( \sin \theta \)

Substituting the value of \( \sin \theta \):
\[ \sin 3\theta = 3 \left( \frac{1}{2} \left( a + \frac{1}{a} \right) \right) - 4 \left( \frac{1}{2} \left( a + \frac{1}{a} \right) \right)^3 \]

### Step 3: Simplify the first term

Calculating the first term:
\[ 3 \cdot \frac{1}{2} \left( a + \frac{1}{a} \right) = \frac{3}{2} \left( a + \frac{1}{a} \right) \]

### Step 4: Calculate \( \sin^3 \theta \)

Now, we need to compute \( \left( \frac{1}{2} \left( a + \frac{1}{a} \right) \right)^3 \):
\[ \left( \frac{1}{2} \left( a + \frac{1}{a} \right) \right)^3 = \frac{1}{8} \left( a + \frac{1}{a} \right)^3 \]

Using the identity \( (x + y)^3 = x^3 + y^3 + 3xy(x + y) \):
\[ \left( a + \frac{1}{a} \right)^3 = a^3 + \frac{1}{a^3} + 3 \left( a \cdot \frac{1}{a} \right) \left( a + \frac{1}{a} \right) = a^3 + \frac{1}{a^3} + 3 \left( a + \frac{1}{a} \right) \]

Thus,
\[ \left( \frac{1}{2} \left( a + \frac{1}{a} \right) \right)^3 = \frac{1}{8} \left( a^3 + \frac{1}{a^3} + 3 \left( a + \frac{1}{a} \right) \right) \]

### Step 5: Substitute back into \( \sin 3\theta \)

Now substituting back into the equation for \( \sin 3\theta \):
\[ \sin 3\theta = \frac{3}{2} \left( a + \frac{1}{a} \right) - 4 \cdot \frac{1}{8} \left( a^3 + \frac{1}{a^3} + 3 \left( a + \frac{1}{a} \right) \right) \]
\[ = \frac{3}{2} \left( a + \frac{1}{a} \right) - \frac{1}{2} \left( a^3 + \frac{1}{a^3} + 3 \left( a + \frac{1}{a} \right) \right) \]

### Step 6: Combine like terms

Combining the terms:
\[ \sin 3\theta = \left( \frac{3}{2} - \frac{3}{2} \right) \left( a + \frac{1}{a} \right) - \frac{1}{2} \left( a^3 + \frac{1}{a^3} \right) \]
\[ = -\frac{1}{2} \left( a^3 + \frac{1}{a^3} \right) \]

### Step 7: Set equal to the given \( \sin 3\theta \)

Now we equate this to the given expression for \( \sin 3\theta \):
\[ -\frac{1}{2} \left( a^3 + \frac{1}{a^3} \right) = \frac{k}{2} \left( a^3 + \frac{1}{a^3} \right) \]

### Step 8: Solve for \( k \)

From this, we can see:
\[ -k = -1 \implies k = 1 \]

### Step 9: Find \( k + 6 \)

Finally, we need to find \( k + 6 \):
\[ k + 6 = 1 + 6 = 7 \]

Thus, the final answer is:
\[ \boxed{7} \]