Chapter 12, Problem 13PT

To determine

To find: Value of $\sec \left(-\frac{9\pi }{4}\right)$ .

Answer to Problem 13PT

  $\sec \left(-\frac{9\pi }{4}\right)=\sqrt{2}$

Explanation of Solution

Given information:

  $\sec \left(-\frac{9\pi }{4}\right)$

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

  $\sec \left(-\frac{9\pi }{4}\right)$

  $\begin{array}{l}2\pi ={360}^0\\ \Rightarrow \frac{2\pi }{2}=\frac{{360}^0}{2}\\ \Rightarrow \pi ={180}^0\\ \Rightarrow -\frac{9}{4}\times \pi =-\frac{9}{4}\times {180}^0\\ \Rightarrow -\frac{9\pi }{4}=-9\times {45}^0\\ \Rightarrow -\frac{9\pi }{4}=-{405}^0\end{array}$

Here $\theta =-{405}^0$

Coterminal angle $={360}^0-{405}^0=-{45}^0$

Again, finding coterminal angle

We get

Coterminal angle $={360}^0-{45}^0={315}^0$

It means

The terminal side of ${315}^0$ lies in Quadrant − IV

Therefore,

  $\begin{array}{l}{\theta }^{\text{'}}={360}^0-\theta \\ \Rightarrow {\theta }^{\text{'}}={360}^0-{315}^0\\ \Rightarrow {\theta }^{\text{'}}={45}^0\end{array}$

Therefore,

  $\sec \left(-\frac{9\pi }{4}\right)=\sec \left({45}^0\right)$

Because Secant function is positive in Quadrant − IV

  $\Rightarrow \sec \left(-\frac{9\pi }{4}\right)=\sqrt{2}$

Therefore,

  $\sec \left(-\frac{9\pi }{4}\right)=\sqrt{2}$