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 step by step, we start with the given equations:

1. **Given:**
\[ \sin \theta = \frac{1}{2} \left( a + \frac{1}{a} \right) \]
\[ \sin 3\theta = \frac{k}{2} \left( a^3 + \frac{1}{a^3} \right) \]

2. **Assume a value for \(\theta\):**
Let's assume \(\theta = 30^\circ\). We know that:
\[ \sin 30^\circ = \frac{1}{2} \]
Therefore, we have:
\[ \frac{1}{2} = \frac{1}{2} \left( a + \frac{1}{a} \right) \]

3. **Simplifying the equation:**
Multiplying both sides by 2:
\[ 1 = a + \frac{1}{a} \]

4. **Rearranging the equation:**
This can be rearranged to:
\[ a + \frac{1}{a} = 1 \]

5. **Cubing both sides:**
We can cube both sides to find \(a^3 + \frac{1}{a^3}\):
\[ \left( a + \frac{1}{a} \right)^3 = a^3 + \frac{1}{a^3} + 3 \left( a + \frac{1}{a} \right) \]
Substituting \(a + \frac{1}{a} = 1\):
\[ 1^3 = a^3 + \frac{1}{a^3} + 3 \cdot 1 \]
\[ 1 = a^3 + \frac{1}{a^3} + 3 \]

6. **Solving for \(a^3 + \frac{1}{a^3}\):**
Rearranging gives:
\[ a^3 + \frac{1}{a^3} = 1 - 3 = -2 \]

7. **Substituting into the second equation:**
Now substituting this value into the equation for \(\sin 3\theta\):
\[ \sin 3\theta = \frac{k}{2} \left( -2 \right) \]
This simplifies to:
\[ \sin 3\theta = -k \]

8. **Finding \(\sin 3\theta\):**
Since \(\theta = 30^\circ\), we have:
\[ 3\theta = 90^\circ \quad \Rightarrow \quad \sin 90^\circ = 1 \]
Thus:
\[ 1 = -k \quad \Rightarrow \quad k = -1 \]

9. **Calculating \(k + 6\):**
Finally, we compute:
\[ k + 6 = -1 + 6 = 5 \]

Therefore, the final answer is:
\[ \boxed{5} \]