Evaluate
(i) `sin [(pi/3 - sin^(-1)(-1/2)]`

Answer

To evaluate \( \sin \left( \frac{\pi}{3} - \sin^{-1} \left( -\frac{1}{2} \right) \right) \), follow these steps:

### Step 1: Simplify the inverse sine expression
First, we use the property of the inverse sine function:
\[ \sin^{-1}(-x) = -\sin^{-1}(x) \]

So,
\[ \sin^{-1} \left( -\frac{1}{2} \right) = -\sin^{-1} \left( \frac{1}{2} \right) \]

### Step 2: Substitute back into the original expression
Now substitute this back into the original expression:
\[ \sin \left( \frac{\pi}{3} - \left( -\sin^{-1} \left( \frac{1}{2} \right) \right) \right) \]
This simplifies to:
\[ \sin \left( \frac{\pi}{3} + \sin^{-1} \left( \frac{1}{2} \right) \right) \]

### Step 3: Determine the value of \( \sin^{-1} \left( \frac{1}{2} \right) \)
We know that:
\[ \sin \left( \frac{\pi}{6} \right) = \frac{1}{2} \]
Thus,
\[ \sin^{-1} \left( \frac{1}{2} \right) = \frac{\pi}{6} \]

### Step 4: Substitute and simplify
Substitute \( \frac{\pi}{6} \) into the expression:
\[ \sin \left( \frac{\pi}{3} + \frac{\pi}{6} \right) \]

Combine the fractions:
\[ \frac{\pi}{3} + \frac{\pi}{6} = \frac{2\pi}{6} + \frac{\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} \]

### Step 5: Evaluate the sine function
Finally, evaluate the sine function:
\[ \sin \left( \frac{\pi}{2} \right) = 1 \]

So, the value is:
\[ \boxed{1} \]