If `0 le x le 1 then cos^(-1)(2x^(2)-1)` equals

Options

A. `2cos^(-1)x`
B. `2sin^(-1)x`
C. `pi-2cos^(-1)x`
D. `pi+2cos^(-1)x`

Answer

To solve the problem \( \cos^{-1}(2x^2 - 1) \) for \( 0 \leq x \leq 1 \), we can follow these steps:

### Step 1: Recognize the identity
We start with the expression \( 2x^2 - 1 \). We can relate this to the cosine double angle identity:
\[ \cos(2\theta) = 2\cos^2(\theta) - 1 \]
This means that if we let \( x = \cos(\theta) \), then:
\[ 2x^2 - 1 = \cos(2\theta) \]

### Step 2: Substitute \( x \) in terms of \( \theta \)
From the substitution \( x = \cos(\theta) \), we can express \( 2x^2 - 1 \) as:
\[ 2x^2 - 1 = \cos(2\theta) \]

### Step 3: Apply the inverse cosine function
Now, we apply the inverse cosine function:
\[ \cos^{-1}(2x^2 - 1) = \cos^{-1}(\cos(2\theta)) \]

### Step 4: Simplify the expression
Since \( \cos^{-1}(\cos(2\theta)) \) will yield \( 2\theta \) when \( 2\theta \) is in the range of \( [0, \pi] \), we have:
\[ \cos^{-1}(2x^2 - 1) = 2\theta \]

### Step 5: Substitute back for \( \theta \)
Recall that \( \theta = \cos^{-1}(x) \), so we substitute back:
\[ 2\theta = 2\cos^{-1}(x) \]

### Final Result
Thus, we conclude that:
\[ \cos^{-1}(2x^2 - 1) = 2\cos^{-1}(x) \]