Find : `cos[2sin^(-1)√((1-x)/2) ]=`

Options

A. `1`
B. `(1)/(4)`
C. `(1)/(2)`
D. `16`

Answer

To solve the problem \( \cos[2 \sin^{-1}(\sqrt{\frac{1-x}{2}})] \), we will follow these steps:

### Step 1: Let \( \alpha = \sin^{-1}(\sqrt{\frac{1-x}{2}}) \)
We can rewrite the expression as:
\[ \cos(2\alpha) \]

### Step 2: Use the double angle formula for cosine
The double angle formula for cosine is:
\[ \cos(2\alpha) = \cos^2(\alpha) - \sin^2(\alpha) \]

### Step 3: Find \( \sin(\alpha) \) and \( \cos(\alpha) \)
From our definition of \( \alpha \):
\[ \sin(\alpha) = \sqrt{\frac{1-x}{2}} \]

To find \( \cos(\alpha) \), we use the Pythagorean identity:
\[ \cos^2(\alpha) + \sin^2(\alpha) = 1 \]
Substituting \( \sin(\alpha) \):
\[ \cos^2(\alpha) + \left(\sqrt{\frac{1-x}{2}}\right)^2 = 1 \]
\[ \cos^2(\alpha) + \frac{1-x}{2} = 1 \]
\[ \cos^2(\alpha) = 1 - \frac{1-x}{2} \]
\[ \cos^2(\alpha) = \frac{2}{2} - \frac{1-x}{2} = \frac{2 - (1-x)}{2} = \frac{1+x}{2} \]

### Step 4: Substitute \( \sin(\alpha) \) and \( \cos(\alpha) \) into the double angle formula
Now substituting back into the double angle formula:
\[ \cos(2\alpha) = \cos^2(\alpha) - \sin^2(\alpha) \]
\[ = \frac{1+x}{2} - \left(\sqrt{\frac{1-x}{2}}\right)^2 \]
\[ = \frac{1+x}{2} - \frac{1-x}{2} \]
\[ = \frac{(1+x) - (1-x)}{2} = \frac{1+x - 1 + x}{2} = \frac{2x}{2} = x \]

### Final Result
Thus, we have:
\[ \cos[2 \sin^{-1}(\sqrt{\frac{1-x}{2}})] = x \]