Express in terms of : `cos^(-1)(2x^(2)-1)` to `cos^(-1)x` for `-1lexlt0`

Explanation

To express \( \cos^{-1}(2x^2 - 1) \) in terms of \( \cos^{-1}(x) \) for \( -1 < x < 0 \), we can follow these steps:

### Step-by-Step Solution:

1. **Substitution**: Let \( x = \cos(\theta) \).
- Since \( x \) is in the interval \( -1 < x < 0 \), this implies \( \theta \) is in the interval \( \frac{\pi}{2} < \theta < \pi \).

**Hint**: Use the relationship between cosine and angles to express \( x \) in terms of \( \theta \).

2. **Rewrite the expression**: We need to express \( 2x^2 - 1 \) in terms of \( \theta \):
\[ 2x^2 - 1 = 2(\cos(\theta))^2 - 1 \]

3. **Use the double angle identity**: Recall the double angle formula for cosine:
\[ 2\cos^2(\theta) - 1 = \cos(2\theta) \]
Therefore,
\[ 2x^2 - 1 = \cos(2\theta) \]

**Hint**: Remember the trigonometric identity that connects cosine of double angles to squares of cosine.

4. **Apply the inverse cosine**: Now we can write:
\[ \cos^{-1}(2x^2 - 1) = \cos^{-1}(\cos(2\theta)) \]

5. **Determine the range**: Since \( \theta \) is in the interval \( \frac{\pi}{2} < \theta < \pi \), it follows that \( 2\theta \) will be in the interval \( \pi < 2\theta < 2\pi \).

6. **Use the property of inverse cosine**: The property of the inverse cosine function states that:
\[ \cos^{-1}(\cos(x)) = x \quad \text{for } x \in [0, \pi] \]
However, since \( 2\theta \) is in the interval \( (\pi, 2\pi) \), we can use:
\[ \cos^{-1}(\cos(2\theta)) = 2\pi - 2\theta \]

7. **Substituting back for \( \theta \)**: Recall that \( \theta = \cos^{-1}(x) \):
\[ \cos^{-1}(2x^2 - 1) = 2\pi - 2\cos^{-1}(x) \]

### Final Answer:
Thus, we can express \( \cos^{-1}(2x^2 - 1) \) in terms of \( \cos^{-1}(x) \) as:
\[ \cos^{-1}(2x^2 - 1) = 2\pi - 2\cos^{-1}(x) \]