The value of `cos(2cos^(-1)x+sin^(-1)x)at x=1//5` is

Options

A. `-(2sqrt6)/5`
B. `(2sqrt5)/5`
C. `(3sqrt5)/5`
D. none

Answer

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

### Step 1: Identify the expression
We need to evaluate:
\[ \cos(2 \cos^{-1} x + \sin^{-1} x) \]
at \( x = \frac{1}{5} \).

### Step 2: Use the identity for inverse functions
We know that:
\[ \cos^{-1} x + \sin^{-1} x = \frac{\pi}{2} \]
Thus, we can rewrite our expression:
\[ \cos(2 \cos^{-1} x + \sin^{-1} x) = \cos(2 \cos^{-1} x + \frac{\pi}{2}) \]

### Step 3: Apply the cosine addition formula
Using the cosine addition formula:
\[ \cos(a + b) = \cos a \cos b - \sin a \sin b \]
we can substitute \( a = 2 \cos^{-1} x \) and \( b = \frac{\pi}{2} \):
\[ \cos(2 \cos^{-1} x + \frac{\pi}{2}) = \cos(2 \cos^{-1} x) \cos\left(\frac{\pi}{2}\right) - \sin(2 \cos^{-1} x) \sin\left(\frac{\pi}{2}\right) \]
Since \( \cos\left(\frac{\pi}{2}\right) = 0 \) and \( \sin\left(\frac{\pi}{2}\right) = 1 \), this simplifies to:
\[ -\sin(2 \cos^{-1} x) \]

### Step 4: Use the double angle formula for sine
We can use the double angle formula:
\[ \sin(2\theta) = 2 \sin \theta \cos \theta \]
where \( \theta = \cos^{-1} x \). Therefore:
\[ \sin(2 \cos^{-1} x) = 2 \sin(\cos^{-1} x) \cos(\cos^{-1} x) \]

### Step 5: Find \( \sin(\cos^{-1} x) \) and \( \cos(\cos^{-1} x) \)
From the definition of inverse cosine:
\[ \cos(\cos^{-1} x) = x \]
To find \( \sin(\cos^{-1} x) \), we use the identity:
\[ \sin(\theta) = \sqrt{1 - \cos^2(\theta)} = \sqrt{1 - x^2} \]
Thus:
\[ \sin(\cos^{-1} x) = \sqrt{1 - x^2} \]

### Step 6: Substitute back into the sine double angle formula
Now substituting these values back:
\[ \sin(2 \cos^{-1} x) = 2 \sqrt{1 - x^2} \cdot x \]
Therefore:
\[ -\sin(2 \cos^{-1} x) = -2x\sqrt{1 - x^2} \]

### Step 7: Substitute \( x = \frac{1}{5} \)
Now we substitute \( x = \frac{1}{5} \):
\[ -\sin(2 \cos^{-1} \frac{1}{5}) = -2 \cdot \frac{1}{5} \cdot \sqrt{1 - \left(\frac{1}{5}\right)^2} \]
Calculating \( 1 - \left(\frac{1}{5}\right)^2 \):
\[ 1 - \frac{1}{25} = \frac{24}{25} \]
Thus:
\[ \sqrt{1 - \left(\frac{1}{5}\right)^2} = \sqrt{\frac{24}{25}} = \frac{\sqrt{24}}{5} = \frac{2\sqrt{6}}{5} \]
Now substituting back:
\[ -2 \cdot \frac{1}{5} \cdot \frac{2\sqrt{6}}{5} = -\frac{4\sqrt{6}}{25} \]

### Final Answer
Thus, the value of \( \cos(2 \cos^{-1} x + \sin^{-1} x) \) at \( x = \frac{1}{5} \) is:
\[ -\frac{4\sqrt{6}}{25} \]