Question

let $\theta=\frac{16pi}{3}$. complete parts (a), (b), and (c) below.
(a) sketch $\theta$ in standard position.

Explanation:

Step1: Simplify the angle

First, find an equivalent angle between \(0\) and \(2\pi\). We know that \(\frac{16\pi}{3}= 4\pi+\frac{4\pi}{3}\). Since \(4\pi\) represents two full - rotations (\(2\times2\pi\)), we can focus on the non - full - rotation part \(\frac{4\pi}{3}\).

Step2: Determine the quadrant

The angle \(\theta=\frac{4\pi}{3}\) is in the third quadrant. The standard position of an angle starts with the initial side on the positive \(x\) - axis. To sketch \(\frac{4\pi}{3}\), we rotate counter - clockwise from the positive \(x\) - axis. The angle \(\frac{\pi}{3}\) (or \(60^{\circ}\)) past \(\pi\) (or \(180^{\circ}\)) is \(\frac{4\pi}{3}\) (or \(240^{\circ}\)).

Answer:

To sketch \(\theta = \frac{16\pi}{3}\) in standard position, start with the initial side on the positive \(x\) - axis. Since \(\frac{16\pi}{3}=4\pi+\frac{4\pi}{3}\), it is equivalent to an angle of \(\frac{4\pi}{3}\) after two full counter - clockwise rotations. Rotate counter - clockwise from the positive \(x\) - axis by an angle of \(\frac{4\pi}{3}\) (or \(240^{\circ}\)), and the terminal side of the angle will be in the third quadrant, making an angle of \(\frac{\pi}{3}\) (or \(60^{\circ}\)) with the negative \(x\) - axis.