Question

The value of \(\cos^{-1}(\cos(\frac{7\pi}{6}))\) is equal to

• A. \(\frac{7\pi}{6}\)
• B. \(\frac{\pi}{6}\)
• C. \(\frac{\pi}{3}\)
• D. \(\frac{2\pi}{3}\)
• E. \(\frac{5\pi}{6}\)

Answer 1

Answer: (E)

We are given the expression \( \cos^{-1} \left( \cos \left( \frac{7\pi}{6} \right) \right) \). The range of the inverse cosine function, \( \cos^{-1} x \), is \( [0, \pi] \). However, \( \frac{7\pi}{6} \) is outside this range. To find the value, we first calculate the cosine of \( \frac{7\pi}{6} \). We know that: \[ \cos \left( \frac{7\pi}{6} \right) = \cos \left( \pi + \frac{\pi}{6} \right) = -\cos \left( \frac{\pi}{6} \right) = -\frac{\sqrt{3}}{2} \] Now, we want to find the angle in the range \( [0, \pi] \) whose cosine is \( -\frac{\sqrt{3}}{2} \). This angle is \( \frac{5\pi}{6} \). Thus, the value of \( \cos^{-1} \left( \cos \left( \frac{7\pi}{6} \right) \right) \) is \( \frac{5\pi}{6} \).

The correct option is (D) : \(\frac{5\pi}{6}\)

Answer 2

We want to find the value of \(\cos^{-1}\left(\cos\left(\frac{7\pi}{6}\right)\right)\).

The range of the inverse cosine function, \(\cos^{-1}(x)\), is \([0, \pi]\).

Since \(\frac{7\pi}{6}\) is not in the range \([0, \pi]\), we cannot directly say that \(\cos^{-1}\left(\cos\left(\frac{7\pi}{6}\right)\right) = \frac{7\pi}{6}\).

We have \(\cos\left(\frac{7\pi}{6}\right) = \cos\left(\pi + \frac{\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}\).

Now we want to find \(\theta\) such that \(\cos\theta = -\frac{\sqrt{3}}{2}\) and \(0 \le \theta \le \pi\).

We know that \(\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}\) and \(\frac{5\pi}{6}\) is in the interval \([0, \pi]\).

Therefore, \(\cos^{-1}\left(\cos\left(\frac{7\pi}{6}\right)\right) = \frac{5\pi}{6}\).