`cos(2cos^(-1)x+sin^(-1)x) " at " x =1/5` is equal to……….. Where `0 le cos^(-1) x le pi and -pi/2 le sin^(-1) x le pi/2`

Answer

To solve the problem \( \cos(2 \cos^{-1} x + \sin^{-1} x) \) at \( x = \frac{1}{5} \), we will follow these steps:

### Step 1: Substitute \( x \)
We start by substituting \( x = \frac{1}{5} \) into the expression:
\[ \cos(2 \cos^{-1}(\frac{1}{5}) + \sin^{-1}(\frac{1}{5})) \]

### Step 2: Define \( \alpha \)
Let \( \alpha = \cos^{-1}(\frac{1}{5}) \). Then we have:
\[ \cos(2\alpha + \sin^{-1}(\frac{1}{5})) \]

### Step 3: Use the identity for \( \cos(2\alpha) \)
Using the double angle identity, we can express \( \cos(2\alpha) \) as:
\[ \cos(2\alpha) = 2 \cos^2(\alpha) - 1 \]
Since \( \cos(\alpha) = \frac{1}{5} \), we find:
\[ \cos^2(\alpha) = \left(\frac{1}{5}\right)^2 = \frac{1}{25} \]
Thus,
\[ \cos(2\alpha) = 2 \cdot \frac{1}{25} - 1 = \frac{2}{25} - 1 = \frac{2 - 25}{25} = \frac{-23}{25} \]

### Step 4: Find \( \sin(\alpha) \)
Next, we need to find \( \sin(\alpha) \). Using the Pythagorean identity:
\[ \sin^2(\alpha) + \cos^2(\alpha) = 1 \]
we have:
\[ \sin^2(\alpha) = 1 - \cos^2(\alpha) = 1 - \frac{1}{25} = \frac{24}{25} \]
Thus,
\[ \sin(\alpha) = \sqrt{\frac{24}{25}} = \frac{\sqrt{24}}{5} = \frac{2\sqrt{6}}{5} \]

### Step 5: Use the identity for \( \cos(2\alpha + \beta) \)
Now we can use the cosine addition formula:
\[ \cos(2\alpha + \sin^{-1}(\frac{1}{5})) = \cos(2\alpha) \cos(\sin^{-1}(\frac{1}{5})) - \sin(2\alpha) \sin(\sin^{-1}(\frac{1}{5})) \]

### Step 6: Find \( \cos(\sin^{-1}(\frac{1}{5})) \)
From the definition of sine and cosine:
\[ \sin(\beta) = \frac{1}{5} \implies \cos(\beta) = \sqrt{1 - \sin^2(\beta)} = \sqrt{1 - \left(\frac{1}{5}\right)^2} = \sqrt{1 - \frac{1}{25}} = \sqrt{\frac{24}{25}} = \frac{2\sqrt{6}}{5} \]

### Step 7: Find \( \sin(2\alpha) \)
Using the double angle formula for sine:
\[ \sin(2\alpha) = 2 \sin(\alpha) \cos(\alpha) = 2 \cdot \frac{2\sqrt{6}}{5} \cdot \frac{1}{5} = \frac{4\sqrt{6}}{25} \]

### Step 8: Substitute back into the cosine formula
Now substituting everything back:
\[ \cos(2\alpha + \sin^{-1}(\frac{1}{5})) = \frac{-23}{25} \cdot \frac{2\sqrt{6}}{5} - \frac{4\sqrt{6}}{25} \cdot \frac{1}{5} \]
Calculating this gives:
\[ = \frac{-46\sqrt{6}}{125} - \frac{4\sqrt{6}}{125} = \frac{-50\sqrt{6}}{125} = \frac{-2\sqrt{6}}{5} \]

### Final Answer
Thus, the final answer is:
\[ \cos(2 \cos^{-1}(\frac{1}{5}) + \sin^{-1}(\frac{1}{5})) = \frac{-2\sqrt{6}}{5} \]