Question

find an angle between 0 and 2π that is coterminal with the given angle. -\frac{4π}{3}

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find an angle between 0 and 2π that is coterminal with the given angle. 8

Explanation:

Step1: Recall coterminal - angle formula

Coterminal angles are found by adding or subtracting \(2k\pi\) (\(k\in\mathbb{Z}\)) to the given angle. For \(\theta =-\frac{4\pi}{3}\), we want to find \(k\) such that \(0\leqslant-\frac{4\pi}{3}+ 2k\pi\leqslant2\pi\).
Let's start with \(k = 1\). Then \(-\frac{4\pi}{3}+2\pi=-\frac{4\pi}{3}+\frac{6\pi}{3}=\frac{-4 + 6}{3}\pi=\frac{2\pi}{3}\). Since \(0<\frac{2\pi}{3}<2\pi\), this is the coterminal - angle for \(-\frac{4\pi}{3}\) in the desired range.

Step2: For the angle \(\theta = 8\) (in radians)

We know that \(2\pi\approx6.28\). Let's find \(k\) such that \(0\leqslant8 - 2k\pi\leqslant2\pi\).
First, solve the left - hand inequality \(0\leqslant8 - 2k\pi\), which gives \(2k\pi\leqslant8\) or \(k\leqslant\frac{4}{\pi}\approx1.27\).
Then, solve the right - hand inequality \(8 - 2k\pi\leqslant2\pi\). Rearranging, we get \(8\leqslant2\pi+2k\pi=(2 + 2k)\pi\), then \(\frac{8}{\pi}\leqslant2 + 2k\), \(\frac{8}{\pi}-2\leqslant2k\), \(k\geqslant\frac{4}{\pi}-1\approx1.27 - 1 = 0.27\).
Since \(k\in\mathbb{Z}\), \(k = 1\). Then \(8-2\pi\approx8 - 6.28 = 1.72\), and \(0<8 - 2\pi<2\pi\).

Answer:

\(\frac{2\pi}{3}\)
\(8 - 2\pi\)