Chapter 7, Problem 1CT

To determine

Find the value of ${\sec }^{-1}\frac{2}{\sqrt{3}}$ . Express angles in radians .

Answer to Problem 1CT

  ${\sec }^{-1}\frac{2}{\sqrt{3}}=\frac{\pi }{6}$

Explanation of Solution

Given information:

Find the value of ${\sec }^{-1}\frac{2}{\sqrt{3}}$ Express angles in radians.

Calculation:

Let $\theta ={\sec }^{-1}\frac{2}{\sqrt{3}}$

Rewrite the inverse function as $\sec \theta =\left(\frac{2}{\sqrt{3}}\right)$

Use the reciprocal identity, $\sec \theta =\frac{1}{\cos \theta }$

  $\cos \theta =\left(\frac{\sqrt{3}}{2}\right)$

Determine the angle for which the cosine function takes the value $\frac{\sqrt{3}}{2}$ for $0\le \theta \le \pi$

  $\boxed{\begin{array}{l}\theta 0\frac{\pi }{6}\frac{\pi }{4}\frac{\pi }{3}\frac{\pi }{2}\frac{2\pi }{3}\frac{3\pi }{4}\frac{5\pi }{6}\pi \\ \cos \theta 1\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}\frac{1}{2}0\frac{-1}{2}\frac{-\sqrt{2}}{2}\frac{-\sqrt{3}}{2}-1\end{array}}$

From the table we see that only angle $\theta$ for which the cosine equals $\frac{\sqrt{3}}{2}$ within theinterval$\left[0,\pi \right]$ is $\frac{\pi }{6}$

So, $\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$ and $\frac{\pi }{6}$ is in the interval $0\le \theta \le \pi$

Hence the value of ${\sec }^{-1}\frac{2}{\sqrt{3}}=\frac{\pi }{6}$