Question

find the reference angle, the quadrant of the terminal side, and the sine and cosine of the angle.
\\( \frac { 7 \pi } { 6 } \\)
reference angle:
quadrant:
\\( \sin \left( \frac { 7 \pi } { 6 } \
ight) = \\)
\\( \cos \left( \frac { 7 \pi } { 6 } \
ight) = \\)
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Explanation:

Step1: Determine the quadrant

Since \(\pi<\frac{7\pi}{6}<\frac{3\pi}{2}\), the angle \(\frac{7\pi}{6}\) is in Quadrant III.

Step2: Calculate the reference angle

The formula for the reference angle \(\theta'\) of an angle \(\theta\) in Quadrant III is \(\theta'=\theta - \pi\).
So, \(\theta'=\frac{7\pi}{6}-\pi=\frac{7\pi - 6\pi}{6}=\frac{\pi}{6}\)

Step3: Find \(\sin(\frac{7\pi}{6})\)

We know that \(\sin(\theta)=-\sin(\theta')\) for \(\theta\) in Quadrant III.
Since \(\sin(\frac{\pi}{6})=\frac{1}{2}\), then \(\sin(\frac{7\pi}{6})=-\frac{1}{2}\)

Step4: Find \(\cos(\frac{7\pi}{6})\)

We know that \(\cos(\theta)=-\cos(\theta')\) for \(\theta\) in Quadrant III.
Since \(\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\), then \(\cos(\frac{7\pi}{6})=-\frac{\sqrt{3}}{2}\)

Answer:

Reference angle: \(\frac{\pi}{6}\)
Quadrant: III
\(\sin(\frac{7\pi}{6})=-\frac{1}{2}\)
\(\cos(\frac{7\pi}{6})=-\frac{\sqrt{3}}{2}\)