Question

find the exact value of the trigonometric function at the given real number.
(a) (sec \frac{11 pi}{3})

(b) (csc \frac{11 pi}{3})

(c) (sec left(-\frac{pi}{6}
ight))

Explanation:

Step1: Find the reference angle for \(\frac{11\pi}{3}\)

\(\frac{11\pi}{3}- 2\pi\times1=\frac{11\pi - 6\pi}{3}=\frac{5\pi}{3}\), \(\frac{5\pi}{3}\) is in the fourth - quadrant. The reference angle \(\theta'=2\pi-\frac{5\pi}{3}=\frac{\pi}{3}\)
\(\sec t=\frac{1}{\cos t}\), \(\cos\frac{11\pi}{3}=\cos\frac{\pi}{3}=\frac{1}{2}\)
\(\sec\frac{11\pi}{3}=\frac{1}{\cos\frac{11\pi}{3}} = 2\)

Step2: Find \(\csc\frac{11\pi}{3}\)

\(\csc t=\frac{1}{\sin t}\), \(\sin\frac{11\pi}{3}=-\sin\frac{\pi}{3}=-\frac{\sqrt{3}}{2}\)
\(\csc\frac{11\pi}{3}=\frac{1}{\sin\frac{11\pi}{3}}=-\frac{2\sqrt{3}}{3}\)

Step3: Find \(\sec(-\frac{\pi}{6})\)

Since \(\sec(-t)=\sec t\) (because \(\sec t=\frac{1}{\cos t}\) and \(\cos(-t)=\cos t\))
\(\sec(-\frac{\pi}{6})=\sec\frac{\pi}{6}\), and \(\sec\frac{\pi}{6}=\frac{1}{\cos\frac{\pi}{6}}\), \(\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}\)
\(\sec(-\frac{\pi}{6})=\frac{2\sqrt{3}}{3}\)

Answer:

(a) \(2\)
(b) \(-\frac{2\sqrt{3}}{3}\)
(c) \(\frac{2\sqrt{3}}{3}\)