find the exact value of the trigonometric function at the given real number.\n(a) sin\\(\\frac { 19\\pi } {…

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})