Problem 42 - Precalculus Essentials

Question

Find the exact value of each trigonometric function. Do not use a calculator. $$\sec \left(-\frac{9 \pi}{4}\right)$$

Step-by-Step Solution

Answer

The exact value of \(\sec \left(-\frac{9 \pi}{4}\right)\) is \(\sqrt{2}\).

Step 1: Simplify the Radian Measure

Bring the radian measure into the range of 0 to \(2 \pi\) by adding a multiple of \(2 \pi\). This is possible because trigonometric functions are periodic functions with period \(2 \pi\). Since the given angle is \(-\frac{9 \pi}{4}\), we add \(2 \pi= \frac{8 \pi}{4}\) to get \(-\frac{9 \pi}{4}+ \frac{8 \pi}{4}= -\frac{\pi}{4}\). So, \(\sec \left(-\frac{9 \pi}{4}\right)\) is equivalent to \(\sec(\frac{-\pi}{4})\).

Step 2: Determine the Quadrant and Find the Reference Angle

The angle is negative, indicating a clockwise direction from the initial side, into the fourth quadrant. The reference angle in the fourth quadrant is \(\frac{\pi}{4}\). In the fourth quadrant, cosine function is positive, and since secant is reciprocally related to cosine, secant is also positive in this quadrant.

Step 3: Find the Exact Value of the Secant Function

For an angle of \(\frac{\pi}{4}\) in the unit circle, we know the cosine value is \(\frac{\sqrt{2}}{2}\). Therefore, the secant of the angle, being the reciprocal of cosine, is \(sec \: \theta = \frac{1}{cos \: \theta} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}\). Thus, the exact value of \(\sec \left(-\frac{9 \pi}{4}\right)\) is \(\sqrt{2}\).