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question 13 find the reference angle, the quadrant of the terminal side…

Question

question 13
find the reference angle, the quadrant of the terminal side, and the sine and cosine of the angle.
\\( \frac { 5 \pi } { 3 } \\)
reference angle:
quadrant:
\\( \sin \left( \frac { 5 \pi } { 3 } \
ight) = \\)
\\( \cos \left( \frac { 5 \pi } { 3 } \
ight) = \\)
question help: message instructor

Explanation:

Step1: Determine the quadrant

Since \(2\pi=\frac{6\pi}{3}\) and \(\frac{3\pi}{2}=\frac{9\pi}{6}=\frac{4.5\pi}{3}\), and \(\frac{3\pi}{2}<\frac{5\pi}{3}<2\pi\), the terminal side of the angle \(\frac{5\pi}{3}\) is in Quadrant IV.

Step2: Calculate the reference angle

The formula for the reference angle \(\theta_{r}\) of an angle \(\theta\) in Quadrant IV is \(\theta_{r} = 2\pi-\theta\). So, \(\theta_{r}=2\pi - \frac{5\pi}{3}=\frac{6\pi - 5\pi}{3}=\frac{\pi}{3}\)

Step3: Find \(\sin(\frac{5\pi}{3})\)

We know that \(\sin\theta=-\sin\theta_{r}\) for \(\theta\) in Quadrant IV. Since \(\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}\), then \(\sin(\frac{5\pi}{3})=-\frac{\sqrt{3}}{2}\)

Step4: Find \(\cos(\frac{5\pi}{3})\)

We know that \(\cos\theta=\cos\theta_{r}\) for \(\theta\) in Quadrant IV. Since \(\cos(\frac{\pi}{3})=\frac{1}{2}\), then \(\cos(\frac{5\pi}{3})=\frac{1}{2}\)

Answer:

Reference angle: \(\frac{\pi}{3}\)
Quadrant: IV
\(\sin(\frac{5\pi}{3})=-\frac{\sqrt{3}}{2}\)
\(\cos(\frac{5\pi}{3})=\frac{1}{2}\)