Minimum value of `(sin^(-1)x)^(2)+(cos^(-1)x)^(2)` is greater than

Options

A. `(pi^(2))/4`
B. `(pi^(2))/16`
C. `(3pi^(2))/4`
D. `(3pi^(2))/32`

Explanation

To find the minimum value of the expression \((\sin^{-1} x)^2 + (\cos^{-1} x)^2\), we can follow these steps:

### Step 1: Use the identity for inverse sine and cosine
We know that:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
Let \(y = \sin^{-1} x\). Then, we can express \(\cos^{-1} x\) in terms of \(y\):
\[ \cos^{-1} x = \frac{\pi}{2} - y \]

### Step 2: Substitute into the expression
Now we can substitute \(\cos^{-1} x\) into the original expression:
\[ (\sin^{-1} x)^2 + (\cos^{-1} x)^2 = y^2 + \left(\frac{\pi}{2} - y\right)^2 \]

### Step 3: Expand the expression
Expanding the expression gives:
\[ y^2 + \left(\frac{\pi}{2} - y\right)^2 = y^2 + \left(\frac{\pi^2}{4} - \pi y + y^2\right) \]
Combining like terms:
\[ = 2y^2 - \pi y + \frac{\pi^2}{4} \]

### Step 4: Find the derivative
To find the minimum value, we take the derivative of the function:
\[ f(y) = 2y^2 - \pi y + \frac{\pi^2}{4} \]
The first derivative is:
\[ f'(y) = 4y - \pi \]

### Step 5: Set the derivative to zero
Setting the first derivative equal to zero to find critical points:
\[ 4y - \pi = 0 \implies y = \frac{\pi}{4} \]

### Step 6: Find the second derivative
Now we find the second derivative to confirm that this point is a minimum:
\[ f''(y) = 4 \]
Since \(f''(y) > 0\), the function is concave up, indicating that \(y = \frac{\pi}{4}\) is indeed a minimum.

### Step 7: Calculate the minimum value
Now, substituting \(y = \frac{\pi}{4}\) back into the expression to find the minimum value:
\[ f\left(\frac{\pi}{4}\right) = 2\left(\frac{\pi}{4}\right)^2 - \pi\left(\frac{\pi}{4}\right) + \frac{\pi^2}{4} \]
Calculating this:
\[ = 2\left(\frac{\pi^2}{16}\right) - \frac{\pi^2}{4} + \frac{\pi^2}{4} \]
\[ = \frac{\pi^2}{8} - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8} \]

### Conclusion
Thus, the minimum value of \((\sin^{-1} x)^2 + (\cos^{-1} x)^2\) is \(\frac{\pi^2}{8}\).