The principal value of `cos^(-1)cos ((2pi)/3)+ sin^(-1) sin((2pi)/3)` is

Options

A. `pi`
B. `pi/2`
C. `pi/3`
D. `(4pi)/3`

Answer

To solve the question, we need to find the principal value of \( \cos^{-1}(\cos((2\pi)/3)) + \sin^{-1}(\sin((2\pi)/3)) \).

### Step-by-Step Solution:

1. **Evaluate \( \cos^{-1}(\cos((2\pi)/3)) \)**:
- The function \( \cos^{-1}(x) \) gives the angle whose cosine is \( x \), and it is defined for \( x \) in the range \([-1, 1]\) with the output in the range \([0, \pi]\).
- Since \( \frac{2\pi}{3} \) is in the interval \((0, \pi)\), we can directly apply the property:
\[ \cos^{-1}(\cos(\theta)) = \theta \quad \text{if } \theta \in [0, \pi] \]
- Therefore:
\[ \cos^{-1}(\cos((2\pi)/3)) = \frac{2\pi}{3} \]

2. **Evaluate \( \sin^{-1}(\sin((2\pi)/3)) \)**:
- The function \( \sin^{-1}(x) \) gives the angle whose sine is \( x \), and it is defined for \( x \) in the range \([-1, 1]\) with the output in the range \([- \frac{\pi}{2}, \frac{\pi}{2}]\).
- We can rewrite \( \sin((2\pi)/3) \) using the sine subtraction identity:
\[ \sin((2\pi)/3) = \sin(\pi - (2\pi)/3) = \sin(\pi/3) \]
- Since \( \sin^{-1}(\sin(\theta)) = \theta \) if \( \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] \), we need to find the equivalent angle for \( \frac{2\pi}{3} \) in this range:
\[ \sin^{-1}(\sin((2\pi)/3)) = \sin^{-1}(\sin(\pi/3)) = \frac{\pi}{3} \]

3. **Combine the results**:
- Now we add the two results together:
\[ \cos^{-1}(\cos((2\pi)/3)) + \sin^{-1}(\sin((2\pi)/3)) = \frac{2\pi}{3} + \frac{\pi}{3} \]
- Simplifying this gives:
\[ \frac{2\pi}{3} + \frac{\pi}{3} = \frac{3\pi}{3} = \pi \]

### Final Answer:
Thus, the principal value of \( \cos^{-1}(\cos((2\pi)/3)) + \sin^{-1}(\sin((2\pi)/3)) \) is \( \pi \).