Question

find the exact value of the trigonometric function at the given real number.
(a) sin\\(\frac { 19\pi } { 6 }\\)
\\( - \frac { 1 } { 2 } \\)
(b) csc\\(\frac { 19\pi } { 6 }\\)
(c) cot\\(\frac { 19\pi } { 6 }\\)

Explanation:

Step1: Simplify the angle

We know that the sine function has a period of \(2\pi\). So, \(\frac{19\pi}{6}=2\pi+\frac{7\pi}{6}\). And \(\sin(x + 2\pi)=\sin x\), so \(\sin\frac{19\pi}{6}=\sin\frac{7\pi}{6}\). Also, \(\frac{7\pi}{6}=\pi+\frac{\pi}{6}\), and \(\sin(\pi +\alpha)=-\sin\alpha\) (where \(\alpha=\frac{\pi}{6}\)), so \(\sin\frac{7\pi}{6}=-\sin\frac{\pi}{6}\). Since \(\sin\frac{\pi}{6}=\frac{1}{2}\), then \(\sin\frac{19\pi}{6}=-\frac{1}{2}\) (this is part (a) result which is given).

Step2: Find the value of \(\csc\frac{19\pi}{6}\)

Recall that \(\csc x=\frac{1}{\sin x}\). Since \(\sin\frac{19\pi}{6}=-\frac{1}{2}\), then \(\csc\frac{19\pi}{6}=\frac{1}{\sin\frac{19\pi}{6}}=\frac{1}{-\frac{1}{2}}=- 2\)

Step3: Find the value of \(\cot\frac{19\pi}{6}\)

First, \(\cos\frac{19\pi}{6}\): \(\frac{19\pi}{6}=2\pi+\frac{7\pi}{6}\), \(\cos(x + 2\pi)=\cos x\), so \(\cos\frac{19\pi}{6}=\cos\frac{7\pi}{6}\). And \(\cos(\pi+\alpha)=-\cos\alpha\) (\(\alpha = \frac{\pi}{6}\)), \(\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}\), so \(\cos\frac{7\pi}{6}=-\frac{\sqrt{3}}{2}\).
Since \(\cot x=\frac{\cos x}{\sin x}\), and \(\sin\frac{19\pi}{6}=-\frac{1}{2}\), \(\cos\frac{19\pi}{6}=-\frac{\sqrt{3}}{2}\), then \(\cot\frac{19\pi}{6}=\frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}=\sqrt{3}\)

Answer:

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