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 properties of inverse trigonometric functions.

### Step-by-step Solution:

1. **Use the property of cosine inverse:**
\[ \cos^{-1}(-x) = \pi - \cos^{-1}(x) \]
Applying this property:
\[ \cos^{-1}(-\frac{1}{2}) = \pi - \cos^{-1}(\frac{1}{2}) \]

2. **Find \(\cos^{-1}(\frac{1}{2})\):**
The angle whose cosine is \(\frac{1}{2}\) is:
\[ \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3} \]
Therefore:
\[ \cos^{-1}(-\frac{1}{2}) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]

3. **Use the property of sine inverse:**
\[ \sin^{-1}(-x) = -\sin^{-1}(x) \]
Applying this property:
\[ \sin^{-1}(-\frac{1}{2}) = -\sin^{-1}(\frac{1}{2}) \]

4. **Find \(\sin^{-1}(\frac{1}{2})\):**
The angle whose sine is \(\frac{1}{2}\) is:
\[ \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6} \]
Therefore:
\[ \sin^{-1}(-\frac{1}{2}) = -\frac{\pi}{6} \]

5. **Calculate \(2 \sin^{-1}(-\frac{1}{2})\):**
\[ 2 \sin^{-1}(-\frac{1}{2}) = 2 \times -\frac{\pi}{6} = -\frac{\pi}{3} \]

6. **Calculate \(\cos^{-1}(-1)\):**
The angle whose cosine is \(-1\) is:
\[ \cos^{-1}(-1) = \pi \]

7. **Substitute all values back into the original expression:**
\[ 2 \cos^{-1}(-\frac{1}{2}) - 2 \sin^{-1}(-\frac{1}{2}) - \cos^{-1}(-1) \]
becomes:
\[ 2 \times \frac{2\pi}{3} - \left(-\frac{\pi}{3}\right) - \pi \]

8. **Simplify the expression:**
\[ = \frac{4\pi}{3} + \frac{\pi}{3} - \pi \]
\[ = \frac{4\pi}{3} + \frac{\pi}{3} - \frac{3\pi}{3} \]
\[ = \frac{4\pi + \pi - 3\pi}{3} = \frac{2\pi}{3} \]

### Final Answer:
The value of \( 2 \cos^{-1}(-\frac{1}{2}) - 2 \sin^{-1}(-\frac{1}{2}) - \cos^{-1}(-1) \) is \( \frac{2\pi}{3} \).

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