If `theta in [4 pi, 5pi] then cos^(-1)(cos theta)` equals

Options

A. `-4pi+theta`
B. `5 pi - theta`
C. `4 pi - theta`
D. `theta-5pi`

Answer

To solve the problem, we need to find the value of \( \cos^{-1}(\cos \theta) \) when \( \theta \) is in the range \( [4\pi, 5\pi] \).

### Step-by-Step Solution:

1. **Understanding the Range of \( \cos^{-1} \)**:
The function \( \cos^{-1}(x) \) has a range of \( [0, \pi] \). This means that for any angle \( \theta \), \( \cos^{-1}(\cos \theta) \) will yield a result within this range.

**Hint**: Remember that \( \cos^{-1}(x) \) returns angles in the range from 0 to \( \pi \).

2. **Finding the Equivalent Angle**:
Since \( \theta \) is in the interval \( [4\pi, 5\pi] \), we need to find an equivalent angle for \( \theta \) that lies within the range of \( [0, \pi] \).

The cosine function is periodic with a period of \( 2\pi \). Therefore, we can reduce \( \theta \) by subtracting \( 2\pi \) until it falls within the desired range.

\[ \theta - 4\pi \quad \text{(since \( 4\pi \) is the lower limit)} \]

Thus, we can express \( \theta \) as:
\[ \theta = 4\pi + x \quad \text{where } 0 \leq x < 2\pi \]

3. **Calculating \( \cos \theta \)**:
Now, we can calculate \( \cos \theta \):
\[ \cos \theta = \cos(4\pi + x) = \cos x \]
This is because \( \cos \) is an even function, and \( \cos(4\pi + x) = \cos x \).

**Hint**: Use the periodicity of the cosine function to simplify the angle.

4. **Finding \( \cos^{-1}(\cos \theta) \)**:
Now, we need to find \( \cos^{-1}(\cos \theta) \):
\[ \cos^{-1}(\cos \theta) = \cos^{-1}(\cos(4\pi + x)) = \cos^{-1}(\cos x) \]
Since \( x \) is in the range \( [0, 2\pi) \), we need to determine where \( x \) falls:
- If \( x \in [0, \pi] \), then \( \cos^{-1}(\cos x) = x \).
- If \( x \in (\pi, 2\pi) \), then \( \cos^{-1}(\cos x) = 2\pi - x \).

5. **Determining the Final Result**:
Since \( \theta \) is in the range \( [4\pi, 5\pi] \), we can conclude:
- If \( \theta = 4\pi + x \) where \( x \in [0, 2\pi) \), then:
- For \( x \in [0, \pi] \), \( \cos^{-1}(\cos \theta) = x \).
- For \( x \in (\pi, 2\pi) \), \( \cos^{-1}(\cos \theta) = 2\pi - x \).

However, since \( \theta \) is always greater than \( 4\pi \), we can express it as:
\[ \cos^{-1}(\cos \theta) = \theta - 4\pi \]

Therefore, the final answer is:
\[ \cos^{-1}(\cos \theta) = \theta - 4\pi \]

### Conclusion:
The value of \( \cos^{-1}(\cos \theta) \) when \( \theta \) is in the interval \( [4\pi, 5\pi] \) is \( \theta - 4\pi \).