Question

1. find any coterminal angles to

\\( \frac { 5 \pi } { 3 } \\) if
\\( 2 \pi < \theta < 8 \pi \\)
find any coterminal angles to
\\( 120 ^ { \circ } \\) if
\\( 600 ^ { \circ } < \theta < 1200 ^ { \circ } \\)
find any coterminal angles to
\\( \frac { 11 \pi } { 6 } \\) if
\\( - 12 \pi < \theta < - 8 \pi \\)

Explanation:

Step1: Recall the formula for coterminal angles

Coterminal angles are given by \(\theta=\alpha + 2k\pi\) (for radians) or \(\theta=\alpha + 360^{\circ}k\) (for degrees), where \(k\in\mathbb{Z}\)

For \(\frac{5\pi}{3}\):

Step2: Find \(k\) for radians

Let \(\theta=\frac{5\pi}{3}+2k\pi\). We want \(2\pi<\theta<8\pi\).
Set \(\frac{5\pi}{3}+2k\pi>2\pi\) and \(\frac{5\pi}{3}+2k\pi<8\pi\)
For \(\frac{5\pi}{3}+2k\pi>2\pi\), we have \(2k\pi>2\pi - \frac{5\pi}{3}=\frac{6\pi - 5\pi}{3}=\frac{\pi}{3}\), so \(k>\frac{1}{6}\)
For \(\frac{5\pi}{3}+2k\pi<8\pi\), we have \(2k\pi<8\pi-\frac{5\pi}{3}=\frac{24\pi - 5\pi}{3}=\frac{19\pi}{3}\), so \(k<\frac{19}{6}\approx3.17\)
Since \(k\in\mathbb{Z}\), \(k = 1,2,3\)
When \(k = 1\), \(\theta=\frac{5\pi}{3}+2\pi=\frac{5\pi + 6\pi}{3}=\frac{11\pi}{3}\)
When \(k = 2\), \(\theta=\frac{5\pi}{3}+4\pi=\frac{5\pi+12\pi}{3}=\frac{17\pi}{3}\)
When \(k = 3\), \(\theta=\frac{5\pi}{3}+6\pi=\frac{5\pi + 18\pi}{3}=\frac{23\pi}{3}\)

For \(120^{\circ}\):

Step3: Find \(k\) for degrees

Let \(\theta=120^{\circ}+360^{\circ}k\). We want \(600^{\circ}<\theta<1200^{\circ}\)
Set \(120^{\circ}+360^{\circ}k>600^{\circ}\) and \(120^{\circ}+360^{\circ}k<1200^{\circ}\)
For \(120^{\circ}+360^{\circ}k>600^{\circ}\), \(360^{\circ}k>600^{\circ}-120^{\circ}=480^{\circ}\), so \(k>\frac{480^{\circ}}{360^{\circ}}=\frac{4}{3}\approx1.33\)
For \(120^{\circ}+360^{\circ}k<1200^{\circ}\), \(360^{\circ}k<1200^{\circ}-120^{\circ}=1080^{\circ}\), so \(k < 3\)
Since \(k\in\mathbb{Z}\), \(k = 2\)
When \(k = 2\), \(\theta=120^{\circ}+720^{\circ}=840^{\circ}\)

For \(\frac{11\pi}{6}\):

Step4: Find \(k\) for radians

Let \(\theta=\frac{11\pi}{6}+2k\pi\). We want \(- 12\pi<\theta<-8\pi\)
Set \(\frac{11\pi}{6}+2k\pi>-12\pi\) and \(\frac{11\pi}{6}+2k\pi<-8\pi\)
For \(\frac{11\pi}{6}+2k\pi>-12\pi\), \(2k\pi>-12\pi-\frac{11\pi}{6}=\frac{-72\pi - 11\pi}{6}=-\frac{83\pi}{6}\), so \(k>-\frac{83}{12}\approx - 6.92\)
For \(\frac{11\pi}{6}+2k\pi<-8\pi\), \(2k\pi<-8\pi-\frac{11\pi}{6}=\frac{-48\pi-11\pi}{6}=-\frac{59\pi}{6}\), so \(k<-\frac{59}{12}\approx - 4.92\)
Since \(k\in\mathbb{Z}\), \(k=-5,-6\)
When \(k=-5\), \(\theta=\frac{11\pi}{6}-10\pi=\frac{11\pi - 60\pi}{6}=-\frac{49\pi}{6}\)
When \(k = - 6\), \(\theta=\frac{11\pi}{6}-12\pi=\frac{11\pi-72\pi}{6}=-\frac{61\pi}{6}\)

Answer:

• For \(\frac{5\pi}{3}\) with \(2\pi<\theta<8\pi\): \(\frac{11\pi}{3},\frac{17\pi}{3},\frac{23\pi}{3}\)
• For \(120^{\circ}\) with \(600^{\circ}<\theta<1200^{\circ}\): \(840^{\circ}\)
• For \(\frac{11\pi}{6}\) with \(-12\pi<\theta<-8\pi\): \(-\frac{49\pi}{6},-\frac{61\pi}{6}\)