Evaluate the following :
`sin ^(2) ( cos ^(-1) . 1/2) + cos^(2) ( sin^(-1) . 1/3) ` .

Answer

To evaluate the expression \( \sin^2(\cos^{-1}(1/2)) + \cos^2(\sin^{-1}(1/3)) \), we can follow these steps:

### Step 1: Evaluate \( \cos^{-1}(1/2) \)
We know that \( \cos(\pi/3) = 1/2 \). Therefore, we have:
\[ \cos^{-1}(1/2) = \frac{\pi}{3} \]

**Hint:** Recall the values of cosine for standard angles to find the angle whose cosine is \( 1/2 \).

### Step 2: Find \( \sin^2(\cos^{-1}(1/2)) \)
Using the identity \( \sin^2(x) + \cos^2(x) = 1 \), we can find:
\[ \sin^2(\cos^{-1}(1/2)) = \sin^2\left(\frac{\pi}{3}\right) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]

**Hint:** Use the Pythagorean identity to find the sine of the angle.

### Step 3: Evaluate \( \sin^{-1}(1/3) \)
Let \( \theta = \sin^{-1}(1/3) \). This means:
\[ \sin(\theta) = \frac{1}{3} \]

### Step 4: Find \( \cos(\theta) \)
Using the Pythagorean identity again, we can find \( \cos(\theta) \):
\[ \cos^2(\theta) = 1 - \sin^2(\theta) = 1 - \left(\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9} \]
Thus,
\[ \cos(\theta) = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \]

**Hint:** Remember to apply the Pythagorean theorem to find the adjacent side in a right triangle.

### Step 5: Find \( \cos^2(\sin^{-1}(1/3)) \)
Now we can calculate:
\[ \cos^2(\sin^{-1}(1/3)) = \left(\frac{2\sqrt{2}}{3}\right)^2 = \frac{8}{9} \]

**Hint:** Square the value of cosine you just found.

### Step 6: Combine the results
Now we can sum the two parts:
\[ \sin^2(\cos^{-1}(1/2)) + \cos^2(\sin^{-1}(1/3)) = \frac{3}{4} + \frac{8}{9} \]

### Step 7: Find a common denominator and add
The common denominator for \( 4 \) and \( 9 \) is \( 36 \):
\[ \frac{3}{4} = \frac{27}{36}, \quad \frac{8}{9} = \frac{32}{36} \]
Thus,
\[ \frac{27}{36} + \frac{32}{36} = \frac{59}{36} \]

### Final Answer
The value of the expression is:
\[ \boxed{\frac{59}{36}} \]