Exercise 13 Page 865

We want to find the exact value of $\sec \left(\text{-}\frac{9\pi }{4}\right).$ To do so, let's start by recalling some trigonometric values for special angles.

Trigonometric Values for Special Angles
Sine	Cosine	Tangent	Cosecant	Secant	Cotangent
$\sin \frac{\pi }{6}=\frac{1}{2}$	$\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$	$\tan \frac{\pi }{6}=\frac{\sqrt{3}}{3}$	$\csc \frac{\pi }{6}=2$	$\sec \frac{\pi }{6}=\frac{2\sqrt{3}}{3}$	$\cot \frac{\pi }{6}=\sqrt{3}$
$\sin \frac{\pi }{4}=\frac{\sqrt{2}}{2}$	$\cos \frac{\pi }{4}=\frac{\sqrt{2}}{2}$	$\tan \frac{\pi }{4}=1$	$\csc \frac{\pi }{4}=\sqrt{2}$	$\sec \frac{\pi }{4}=\sqrt{2}$	$\cot \frac{\pi }{4}=1$
$\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$	$\cos \frac{\pi }{3}=\frac{1}{2}$	$\tan \frac{\pi }{3}=\sqrt{3}$	$\csc \frac{\pi }{3}=\frac{2\sqrt{3}}{3}$	$\sec \frac{\pi }{3}=2$	$\cot \frac{\pi }{3}=\frac{\sqrt{3}}{3}$

Next, let's graph $\theta =\text{-}\frac{9\pi }{4}$ in standard position so that we can find its reference angle. This way we can use the values from our table. Note that the terminal side of this angle lies in Quadrant IV. Therefore, to find its reference angle ${\theta }^{\prime },$ we will subtract $2\pi$ from $\frac{9\pi }{4}.$

Let's simplify $\frac{9\pi }{4}-2\pi$ to obtain ${\theta }^{\prime }.$ We found that ${\theta }^{\prime }=\frac{\pi }{4}.$ Next, we will recall the signs of the six trigonometric functions in the different quadrants of the coordinate plane. In Quadrant IV, the quadrant where the terminal side of the angle is located, secant is positive. With this information, we can write an equation relating the secant of the angle and the secant of its reference angle. Using our table, we can see that $\sec \frac{\pi }{4}=\sqrt{2}.$ By the Transitive Property of Equality, we can write the value for $\sec \left(\text{-}\frac{9\pi }{4}\right).$