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let $\theta=\frac{16pi}{7}$. complete parts (a), (b), and (c) below. (a…

Question

let $\theta=\frac{16pi}{7}$. complete parts (a), (b), and (c) below. (a) sketch $\theta$ in standard position. (b) find an angle between 0 and $2pi$ that is coterminal with $\theta$. write your answer in radians in terms of $pi$. radians (c) find an angle between $- 2pi$ and 0 that is coterminal with $\theta$. write your answer in radians in terms of $pi$. radians

Explanation:

Step1: Recall coterminal - angle formula

Coterminal angles of $\theta$ are given by $\theta + 2k\pi$, where $k\in\mathbb{Z}$. Given $\theta=\frac{16\pi}{7}$.

Step2: Find a coterminal angle between 0 and $2\pi$ for part (b)

We want to find $k$ such that $0<\frac{16\pi}{7}+2k\pi < 2\pi$.
First, solve the left - hand inequality $0<\frac{16\pi}{7}+2k\pi$:
\[

$$\begin{align*} 0&<\frac{16\pi}{7}+2k\pi\\ - \frac{16\pi}{7}&<2k\pi\\ -\frac{8}{7}&

\]
Then, solve the right - hand inequality $\frac{16\pi}{7}+2k\pi < 2\pi$:
\[

$$\begin{align*} \frac{16\pi}{7}+2k\pi&<2\pi\\ 2k\pi&<2\pi-\frac{16\pi}{7}\\ 2k\pi&=\frac{14\pi - 16\pi}{7}\\ 2k\pi&=-\frac{2\pi}{7}\\ k&<-\frac{1}{7} \end{align*}$$

\]
Since $k\in\mathbb{Z}$, $k = - 1$.
\[

$$\begin{align*} \theta_{1}&=\frac{16\pi}{7}+2(-1)\pi\\ &=\frac{16\pi}{7}-2\pi\\ &=\frac{16\pi - 14\pi}{7}\\ &=\frac{2\pi}{7} \end{align*}$$

\]

Step3: Find a coterminal angle between $-2\pi$ and 0 for part (c)

We want to find $k$ such that $-2\pi<\frac{16\pi}{7}+2k\pi < 0$.
First, solve the left - hand inequality $-2\pi<\frac{16\pi}{7}+2k\pi$:
\[

$$\begin{align*} -2\pi&<\frac{16\pi}{7}+2k\pi\\ -2\pi-2k\pi&<\frac{16\pi}{7}\\ -2\pi(1 + k)&<\frac{16\pi}{7}\\ 1 + k&>-\frac{8}{7}\\ k&>-\frac{15}{7}\approx - 2.14 \end{align*}$$

\]
Then, solve the right - hand inequality $\frac{16\pi}{7}+2k\pi < 0$:
\[

$$\begin{align*} \frac{16\pi}{7}+2k\pi&<0\\ 2k\pi&<-\frac{16\pi}{7}\\ k&<-\frac{8}{7}\approx - 1.14 \end{align*}$$

\]
Since $k\in\mathbb{Z}$, $k=-2$.
\[

$$\begin{align*} \theta_{2}&=\frac{16\pi}{7}+2(-2)\pi\\ &=\frac{16\pi}{7}-4\pi\\ &=\frac{16\pi - 28\pi}{7}\\ &=-\frac{12\pi}{7} \end{align*}$$

\]

Answer:

(b) $\frac{2\pi}{7}$
(c) $-\frac{12\pi}{7}$