`cos^(-1)(cos"(7pi)/6)=`

Options

A. `(7pi)/6`
B. `(5pi)/6`
C. `pi/3`
D. `pi/6`

Answer

To solve the problem \( \cos^{-1}(\cos(7\pi/6)) \), we will follow these steps:

### Step 1: Understanding the Inverse Function
The expression \( \cos^{-1}(x) \) gives the angle whose cosine is \( x \). The range of \( \cos^{-1}(x) \) is \( [0, \pi] \).

### Step 2: Calculate \( \cos(7\pi/6) \)
First, we need to find the cosine of \( 7\pi/6 \):
\[ \cos(7\pi/6) = \cos\left(\pi + \frac{\pi}{6}\right) \]
Using the cosine addition formula, we know that:
\[ \cos(\pi + x) = -\cos(x) \]
Thus,
\[ \cos(7\pi/6) = -\cos\left(\frac{\pi}{6}\right) \]
Since \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \), we have:
\[ \cos(7\pi/6) = -\frac{\sqrt{3}}{2} \]

### Step 3: Setting Up the Inverse Function
Now we substitute back into the inverse function:
\[ \cos^{-1}(\cos(7\pi/6)) = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) \]

### Step 4: Finding the Angle
We need to find an angle \( \theta \) such that:
\[ \cos(\theta) = -\frac{\sqrt{3}}{2} \]
The angles for which cosine is negative are in the second and third quadrants. The reference angle for \( \frac{\sqrt{3}}{2} \) is \( \frac{\pi}{6} \).

Thus, the angles where \( \cos(\theta) = -\frac{\sqrt{3}}{2} \) are:
\[ \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \quad \text{(in the second quadrant)} \]
and
\[ \theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \quad \text{(in the third quadrant)} \]

### Step 5: Selecting the Correct Angle
Since the range of \( \cos^{-1}(x) \) is \( [0, \pi] \), we only consider the angle in the second quadrant:
\[ \theta = \frac{5\pi}{6} \]

### Final Answer
Thus, the final answer is:
\[ \cos^{-1}(\cos(7\pi/6)) = \frac{5\pi}{6} \]

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