The principal value of `sin^(-1)(-sqrt(3)//2)+cos^(-1)(cos(7pi//6))` is

Options

A. `5pi//6`
B. `pi//2`
C. `3pi//2`
D. None of these

Answer

To solve the expression \( \sin^{-1}(-\frac{\sqrt{3}}{2}) + \cos^{-1}(\cos(\frac{7\pi}{6})) \), we will break it down step by step.

### Step 1: Evaluate \( \sin^{-1}(-\frac{\sqrt{3}}{2}) \)

The value of \( \sin^{-1}(x) \) gives us the angle whose sine is \( x \). Here, we need to find the angle whose sine is \( -\frac{\sqrt{3}}{2} \).

The sine function is negative in the fourth quadrant, and the reference angle for \( \frac{\sqrt{3}}{2} \) is \( \frac{\pi}{3} \). Therefore, the angle in the fourth quadrant is:

\[ \sin^{-1}(-\frac{\sqrt{3}}{2}) = -\frac{\pi}{3} \]

### Step 2: Evaluate \( \cos^{-1}(\cos(\frac{7\pi}{6})) \)

Next, we need to evaluate \( \cos^{-1}(\cos(\frac{7\pi}{6})) \). The cosine function is periodic, and \( \frac{7\pi}{6} \) is in the third quadrant where cosine is negative.

The principal value of \( \cos^{-1}(x) \) is defined in the range \( [0, \pi] \). To find the equivalent angle in this range, we can use the property that:

\[ \cos(\theta) = \cos(2\pi - \theta) \]

Thus, we can find the reference angle:

\[ \frac{7\pi}{6} = \pi + \frac{\pi}{6} \]

The reference angle is \( \frac{\pi}{6} \), so:

\[ \cos^{-1}(\cos(\frac{7\pi}{6})) = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \]

### Step 3: Combine the results

Now we combine the results from Step 1 and Step 2:

\[ \sin^{-1}(-\frac{\sqrt{3}}{2}) + \cos^{-1}(\cos(\frac{7\pi}{6})) = -\frac{\pi}{3} + \frac{5\pi}{6} \]

### Step 4: Find a common denominator

To add these fractions, we need a common denominator. The least common multiple of 3 and 6 is 6.

Converting \( -\frac{\pi}{3} \):

\[ -\frac{\pi}{3} = -\frac{2\pi}{6} \]

Now we can add:

\[ -\frac{2\pi}{6} + \frac{5\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} \]

### Final Answer

Thus, the principal value of \( \sin^{-1}(-\frac{\sqrt{3}}{2}) + \cos^{-1}(\cos(\frac{7\pi}{6})) \) is:

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