Question

for the rotation $-\frac{13pi}{3}$, find the coterminal angle from $0 \leq \theta < 2\pi$, the quadrant, and the reference angle.

answer attempt 1 out of 2

the coterminal angle is \boxed{}, which lies in quadrant \underline{\hspace{1cm}}, with a reference angle of \boxed{}.

Explanation:

Step1: Find coterminal angle

To find a coterminal angle with \(-\frac{13\pi}{3}\) in the range \(0\leq\theta < 2\pi\), we add multiples of \(2\pi\) (since coterminal angles differ by \(2\pi n\), \(n\in\mathbb{Z}\)). Let's find \(n\) such that the angle is in the desired range.

We know that \(2\pi=\frac{6\pi}{3}\). Let's add \(3\) times \(2\pi\) (since \(-\frac{13\pi}{3}+ 3\times2\pi=-\frac{13\pi}{3}+\frac{18\pi}{3}=\frac{5\pi}{3}\)). Let's check: \(-\frac{13\pi}{3}+ 3\times2\pi=-\frac{13\pi}{3}+\frac{18\pi}{3}=\frac{5\pi}{3}\), and \(0\leq\frac{5\pi}{3}<2\pi\) (since \(2\pi = \frac{6\pi}{3}\) and \(\frac{5\pi}{3}<\frac{6\pi}{3}\)). So the coterminal angle is \(\frac{5\pi}{3}\).

Step2: Determine the quadrant

Angles in standard position:

• Quadrant I: \(0 < \theta<\frac{\pi}{2}\)
• Quadrant II: \(\frac{\pi}{2}<\theta < \pi\)
• Quadrant III: \(\pi<\theta <\frac{3\pi}{2}\)
• Quadrant IV: \(\frac{3\pi}{2}<\theta < 2\pi\)

\(\frac{5\pi}{3}\) is between \(\frac{3\pi}{2}=\frac{9\pi}{6}\) and \(2\pi=\frac{12\pi}{6}\)? Wait, \(\frac{3\pi}{2}=\frac{9\pi}{6}\), \(\frac{5\pi}{3}=\frac{10\pi}{6}\), \(2\pi=\frac{12\pi}{6}\). So \(\frac{3\pi}{2}<\frac{5\pi}{3}<2\pi\), so it lies in Quadrant IV.

Step3: Find the reference angle

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

Answer:

The coterminal angle is \(\boldsymbol{\frac{5\pi}{3}}\), which lies in Quadrant \(\boldsymbol{IV}\), with a reference angle of \(\boldsymbol{\frac{\pi}{3}}\).