Let `cos^(-1)(4x^(3)-3x)=a+b cos^(-1)x`
Q. If `x in [-(1)/(2), (1)/(2)]`, then `sin^(-1)("sin"(a)/(b))` is :

Options

A. `-(pi)/(3)`
B. `(pi)/(3)`
C. `-(pi)/(6)`
D. `(pi)/(6)`

Answer

To solve the problem, we need to find the value of \( \sin^{-1}\left(\frac{\sin a}{b}\right) \) given that \( \cos^{-1}(4x^3 - 3x) = a + b \cos^{-1}(x) \) for \( x \in \left[-\frac{1}{2}, \frac{1}{2}\right] \).

### Step-by-Step Solution:

1. **Understanding the Given Equation**:
We start with the equation:
\[ \cos^{-1}(4x^3 - 3x) = a + b \cos^{-1}(x) \]
The expression \( 4x^3 - 3x \) is known to be equal to \( \cos(3\theta) \) if we let \( x = \cos(\theta) \). This is derived from the triple angle formula for cosine.

2. **Substituting \( x = \cos(\theta) \)**:
Substitute \( x = \cos(\theta) \):
\[ \cos^{-1}(4\cos^3(\theta) - 3\cos(\theta)) = a + b \cos^{-1}(\cos(\theta)) \]
This simplifies to:
\[ \cos^{-1}(\cos(3\theta)) = a + b\theta \]
Since \( \cos^{-1}(\cos(3\theta)) = 3\theta \) (for \( \theta \) in the appropriate range), we have:
\[ 3\theta = a + b\theta \]

3. **Rearranging the Equation**:
Rearranging gives:
\[ 3\theta - b\theta = a \implies \theta(3 - b) = a \]
Thus, we can express \( \theta \) as:
\[ \theta = \frac{a}{3 - b} \]

4. **Finding the Values of \( a \) and \( b \)**:
Given that \( x \in \left[-\frac{1}{2}, \frac{1}{2}\right] \), we can evaluate the limits:
- For \( x = \frac{1}{2} \):
\[ \cos^{-1}(4\left(\frac{1}{2}\right)^3 - 3\left(\frac{1}{2}\right)) = \cos^{-1}(4 \cdot \frac{1}{8} - \frac{3}{2}) = \cos^{-1}(0) = \frac{\pi}{2} \]
- For \( x = -\frac{1}{2} \):
\[ \cos^{-1}(4\left(-\frac{1}{2}\right)^3 - 3\left(-\frac{1}{2}\right)) = \cos^{-1}(-1) = \pi \]

5. **Setting Up the Equations**:
From the evaluations:
- At \( x = \frac{1}{2} \):
\[ \frac{\pi}{2} = a + b \cdot \frac{\pi}{3} \]
- At \( x = -\frac{1}{2} \):
\[ \pi = a + b \cdot \frac{2\pi}{3} \]

6. **Solving for \( a \) and \( b \)**:
We now have two equations:
1. \( a + \frac{b\pi}{3} = \frac{\pi}{2} \)
2. \( a + \frac{2b\pi}{3} = \pi \)

Subtract the first from the second:
\[ \left(a + \frac{2b\pi}{3}\right) - \left(a + \frac{b\pi}{3}\right) = \pi - \frac{\pi}{2} \]
This simplifies to:
\[ \frac{b\pi}{3} = \frac{\pi}{2} \implies b = \frac{3}{2} \]

Substitute \( b \) back into the first equation to find \( a \):
\[ a + \frac{3\pi}{6} = \frac{\pi}{2} \implies a = \frac{\pi}{2} - \frac{\pi}{2} = 0 \]

7. **Final Calculation**:
Now, we need to find \( \sin^{-1}\left(\frac{\sin a}{b}\right) \):
\[ \sin^{-1}\left(\frac{\sin(0)}{\frac{3}{2}}\right) = \sin^{-1}(0) = 0 \]

### Conclusion:
The final answer is:
\[ \sin^{-1}\left(\frac{\sin a}{b}\right) = 0 \]