If `x=(1)/(5)`, the value of `cos(cos^(-1)x+2sin^(-1)x)` is :

Options

A. `-sqrt((24)/(25))`
B. `sqrt((24)/(25))`
C. `-(1)/(5)`
D. `(1)/(5)`

Answer

To solve the problem, we need to find the value of \( \cos(\cos^{-1}(x) + 2\sin^{-1}(x)) \) given \( x = \frac{1}{5} \).

### Step-by-Step Solution:

1. **Substituting the value of x**:
Given \( x = \frac{1}{5} \), we need to evaluate:
\[ \cos\left(\cos^{-1}\left(\frac{1}{5}\right) + 2\sin^{-1}\left(\frac{1}{5}\right)\right) \]

2. **Using the identity for cosine of a sum**:
We can use the cosine addition formula:
\[ \cos(A + B) = \cos A \cos B - \sin A \sin B \]
Here, let \( A = \cos^{-1}\left(\frac{1}{5}\right) \) and \( B = 2\sin^{-1}\left(\frac{1}{5}\right) \). Thus, we have:
\[ \cos\left(\cos^{-1}\left(\frac{1}{5}\right) + 2\sin^{-1}\left(\frac{1}{5}\right)\right) = \cos A \cos B - \sin A \sin B \]

3. **Finding \( \cos A \) and \( \sin A \)**:
Since \( A = \cos^{-1}\left(\frac{1}{5}\right) \):
\[ \cos A = \frac{1}{5} \]
To find \( \sin A \), we use the identity \( \sin^2 A + \cos^2 A = 1 \):
\[ \sin^2 A = 1 - \left(\frac{1}{5}\right)^2 = 1 - \frac{1}{25} = \frac{24}{25} \]
Therefore, \( \sin A = \sqrt{\frac{24}{25}} = \frac{\sqrt{24}}{5} = \frac{2\sqrt{6}}{5} \).

4. **Finding \( B = 2\sin^{-1}\left(\frac{1}{5}\right) \)**:
To find \( \cos B \) and \( \sin B \), we use the double angle formulas:
\[ \sin(2\theta) = 2\sin(\theta)\cos(\theta) \]
where \( \theta = \sin^{-1}\left(\frac{1}{5}\right) \). We have:
\[ \sin B = 2\sin\left(\sin^{-1}\left(\frac{1}{5}\right)\right)\cos\left(\sin^{-1}\left(\frac{1}{5}\right)\right) \]
Here, \( \sin\left(\sin^{-1}\left(\frac{1}{5}\right)\right) = \frac{1}{5} \) and \( \cos\left(\sin^{-1}\left(\frac{1}{5}\right)\right) = \sqrt{1 - \left(\frac{1}{5}\right)^2} = \sqrt{\frac{24}{25}} = \frac{2\sqrt{6}}{5} \).
Thus:
\[ \sin B = 2 \cdot \frac{1}{5} \cdot \frac{2\sqrt{6}}{5} = \frac{4\sqrt{6}}{25} \]

For \( \cos B \):
\[ \cos B = \cos(2\theta) = 1 - 2\sin^2(\theta) = 1 - 2\left(\frac{1}{5}\right)^2 = 1 - 2\cdot\frac{1}{25} = 1 - \frac{2}{25} = \frac{23}{25} \]

5. **Substituting back into the cosine addition formula**:
Now substituting \( \cos A \), \( \sin A \), \( \cos B \), and \( \sin B \):
\[ \cos\left(\cos^{-1}\left(\frac{1}{5}\right) + 2\sin^{-1}\left(\frac{1}{5}\right)\right) = \left(\frac{1}{5}\right)\left(\frac{23}{25}\right) - \left(\frac{2\sqrt{6}}{5}\right)\left(\frac{4\sqrt{6}}{25}\right) \]
Simplifying this:
\[ = \frac{23}{125} - \frac{8 \cdot 6}{125} = \frac{23}{125} - \frac{48}{125} = \frac{23 - 48}{125} = \frac{-25}{125} = -\frac{1}{5} \]

Thus, the final answer is:
\[ \cos\left(\cos^{-1}\left(\frac{1}{5}\right) + 2\sin^{-1}\left(\frac{1}{5}\right)\right) = -\frac{1}{5} \]