Find the principal value of `cos^(-1)(-(1)/(2))+2sin^(-1)(-(1)/(2))`

Options

A. `(pi)/(2)`
B. `(2pi)/(3)`
C. `(3pi)/(4)`
D. `(5pi)/(8)`

Answer

To find the principal value of the expression \( \cos^{-1}\left(-\frac{1}{2}\right) + 2\sin^{-1}\left(-\frac{1}{2}\right) \), we will follow these steps:

### Step 1: Evaluate \( \cos^{-1}\left(-\frac{1}{2}\right) \)

The range of \( \cos^{-1}(x) \) is from \( 0 \) to \( \pi \). We need to find an angle \( \theta \) such that \( \cos(\theta) = -\frac{1}{2} \).

The angle that satisfies this in the specified range is:
\[ \theta = \frac{2\pi}{3} \]

Thus,
\[ \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3} \]

### Step 2: Evaluate \( 2\sin^{-1}\left(-\frac{1}{2}\right) \)

The range of \( \sin^{-1}(x) \) is from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \). We need to find an angle \( \phi \) such that \( \sin(\phi) = -\frac{1}{2} \).

The angle that satisfies this is:
\[ \phi = -\frac{\pi}{6} \]

Thus,
\[ \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \]

Now, we multiply this by 2:
\[ 2\sin^{-1}\left(-\frac{1}{2}\right) = 2 \times -\frac{\pi}{6} = -\frac{\pi}{3} \]

### Step 3: Combine the results

Now we can combine the results from Step 1 and Step 2:
\[ \cos^{-1}\left(-\frac{1}{2}\right) + 2\sin^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3} - \frac{\pi}{3} \]

### Step 4: Simplify

Now simplify the expression:
\[ \frac{2\pi}{3} - \frac{\pi}{3} = \frac{2\pi - \pi}{3} = \frac{\pi}{3} \]

### Final Answer

Thus, the principal value of \( \cos^{-1}\left(-\frac{1}{2}\right) + 2\sin^{-1}\left(-\frac{1}{2}\right) \) is:
\[ \frac{\pi}{3} \]

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