Write the principal value of :
`cos^(-1)(1/2) - 2 sin^(-1) (-1/2)`

Answer

To find the principal value of the expression \( \cos^{-1}(1/2) - 2 \sin^{-1}(-1/2) \), we will solve it step by step.

### Step 1: Evaluate \( \cos^{-1}(1/2) \)

The value of \( \cos^{-1}(1/2) \) corresponds to the angle whose cosine is \( 1/2 \). The angle in the range of \( [0, \pi] \) that satisfies this is:

\[ \cos^{-1}(1/2) = \frac{\pi}{3} \]

**Hint:** Remember that \( \cos(\frac{\pi}{3}) = \frac{1}{2} \).

### Step 2: Evaluate \( \sin^{-1}(-1/2) \)

Next, we need to find \( \sin^{-1}(-1/2) \). The value of \( \sin^{-1}(-1/2) \) corresponds to the angle whose sine is \( -1/2 \). The angle in the range of \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) that satisfies this is:

\[ \sin^{-1}(-1/2) = -\frac{\pi}{6} \]

**Hint:** Recall that \( \sin(-\frac{\pi}{6}) = -\frac{1}{2} \).

### Step 3: Substitute into the expression

Now substitute the values we found into the original expression:

\[ \cos^{-1}(1/2) - 2 \sin^{-1}(-1/2) = \frac{\pi}{3} - 2 \left(-\frac{\pi}{6}\right) \]

### Step 4: Simplify the expression

Calculating the second term:

\[ -2 \left(-\frac{\pi}{6}\right) = \frac{\pi}{3} \]

Now substituting back into the expression:

\[ \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3} \]

### Final Answer

Thus, the principal value of the expression \( \cos^{-1}(1/2) - 2 \sin^{-1}(-1/2) \) is:

\[ \frac{2\pi}{3} \]

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