The value of `2cos^(-1)(-1/2)-2 sin^(-1)(-1/2)-cos^(-1)(-1)` is

Options

A. `2pi/3`
B. `pi/2`
C. `pi`
D. `2pi`

Answer

To solve the expression \(2\cos^{-1}(-\frac{1}{2}) - 2\sin^{-1}(-\frac{1}{2}) - \cos^{-1}(-1)\), we will use some properties of inverse trigonometric functions.

### Step-by-Step Solution:

1. **Apply the properties of inverse trigonometric functions:**
- We know that \(\cos^{-1}(-x) = \pi - \cos^{-1}(x)\) and \(\sin^{-1}(-x) = -\sin^{-1}(x)\).
- Therefore, we can rewrite the expression:
\[ 2\cos^{-1}(-\frac{1}{2}) = 2(\pi - \cos^{-1}(\frac{1}{2})) = 2\pi - 2\cos^{-1}(\frac{1}{2}) \]
\[ -2\sin^{-1}(-\frac{1}{2}) = -2(-\sin^{-1}(\frac{1}{2})) = 2\sin^{-1}(\frac{1}{2}) \]
\[ -\cos^{-1}(-1) = -(\pi) \quad \text{(since } \cos^{-1}(-1) = \pi\text{)} \]

Thus, the expression becomes:
\[ 2\pi - 2\cos^{-1}(\frac{1}{2}) + 2\sin^{-1}(\frac{1}{2}) - \pi \]

2. **Simplify the expression:**
\[ = (2\pi - \pi) + 2\sin^{-1}(\frac{1}{2}) - 2\cos^{-1}(\frac{1}{2}) \]
\[ = \pi + 2\sin^{-1}(\frac{1}{2}) - 2\cos^{-1}(\frac{1}{2}) \]

3. **Find the values of \(\sin^{-1}(\frac{1}{2})\) and \(\cos^{-1}(\frac{1}{2})\):**
- \(\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}\)
- \(\cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}\)

4. **Substitute these values into the expression:**
\[ = \pi + 2\left(\frac{\pi}{6}\right) - 2\left(\frac{\pi}{3}\right) \]
\[ = \pi + \frac{\pi}{3} - \frac{2\pi}{3} \]
\[ = \pi + \frac{\pi}{3} - \frac{2\pi}{3} = \pi - \frac{\pi}{3} \]
\[ = \frac{3\pi}{3} - \frac{\pi}{3} = \frac{2\pi}{3} \]

5. **Final Result:**
\[ \text{The value of } 2\cos^{-1}(-\frac{1}{2}) - 2\sin^{-1}(-\frac{1}{2}) - \cos^{-1}(-1) = \frac{2\pi}{3} \]