Chapter 13.3, Problem 31PPS

To determine

To find: Value of $\sec \left(\frac{11\pi }{6}\right)$ .

Answer to Problem 31PPS

  $\sec \left(\frac{11\pi }{6}\right)=\frac{2}{\sqrt{3}}$

Explanation of Solution

Given information:

  $\sec \left(\frac{11\pi }{6}\right)$

Concept:

If $\theta$ is the non-quadrantal angle in standard position, its reference angle ${\theta }^{\text{'}}$ is the acute angle formed by the terminal side of $\theta$ and the x-axis.

The ruled for finding the measures of reference angles for $0^0<\theta <{360}^0$ or $0^0<\theta <2\pi$ are shown

For Quadrant − I

  ${\theta }^{\text{'}}=\theta$

For Quadrant − II

  ${\theta }^{\text{'}}={180}^0-\theta$ or ${\theta }^{\text{'}}=\pi -\theta$

For Quadrant − III

  ${\theta }^{\text{'}}=\theta -{180}^0$ or ${\theta }^{\text{'}}=\theta -\pi$

For Quadrant − IV

  ${\theta }^{\text{'}}={360}^0-\theta$ or ${\theta }^{\text{'}}=2\pi -\theta$

Given

  $\sec \left(\frac{11\pi }{6}\right)$

Here $\theta =\frac{11\pi }{6}$

It means

The terminal side of $\frac{11\pi }{6}$ lies in Quadrant − IV

Therefore,

  $\begin{array}{l}{\theta }^{\text{'}}=2\pi -\theta \\ \Rightarrow {\theta }^{\text{'}}=2\pi -\frac{11\pi }{6}\\ \Rightarrow {\theta }^{\text{'}}=\frac{12\pi -11\pi }{6}\\ \Rightarrow {\theta }^{\text{'}}=\frac{\pi }{6}\end{array}$

Therefore,

  $\sec \left(\frac{11\pi }{6}\right)=\sec \left(\frac{\pi }{6}\right)$

Because the Secant function is positive in Quadrant − IV

  $\Rightarrow \sec \left(\frac{11\pi }{6}\right)=\frac{2}{\sqrt{3}}$

Therefore,

  $\sec \left(\frac{11\pi }{6}\right)=\frac{2}{\sqrt{3}}$