Write the principal value of
`cos^(-1)(sqrt(3)//2)`

Answer

To find the principal value of \( \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) \), we will follow these steps:

### Step 1: Identify the value for which the cosine equals \( \frac{\sqrt{3}}{2} \).
We know from trigonometric values that:
\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]

### Step 2: Write the expression in terms of cosine.
We can express \( \frac{\sqrt{3}}{2} \) as:
\[ \frac{\sqrt{3}}{2} = \cos\left(\frac{\pi}{6}\right) \]

### Step 3: Apply the inverse cosine function.
Now, we apply the inverse cosine function:
\[ \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \cos^{-1}\left(\cos\left(\frac{\pi}{6}\right)\right) \]

### Step 4: Simplify using the property of inverse functions.
Using the property that \( \cos^{-1}(\cos(x)) = x \) for \( x \) in the range of \( [0, \pi] \):
\[ \cos^{-1}\left(\cos\left(\frac{\pi}{6}\right)\right) = \frac{\pi}{6} \]

### Step 5: Check if the result is within the principal value range.
The principal value of the inverse cosine function is defined in the range \( [0, \pi] \). Since \( \frac{\pi}{6} \) is within this range, we conclude:
\[ \text{Principal value} = \frac{\pi}{6} \]

### Final Answer:
Thus, the principal value of \( \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) \) is:
\[ \frac{\pi}{6} \]

---