Question

find the reference angle for the angle -\frac{41\pi}{6}.
the reference angle is
(type your answer in radians. type an integer or a simplified fraction.)

Explanation:

Step1: Make the angle positive

Add \(2k\pi\) (\(k\in\mathbb{Z}\)) to make the angle positive. Let's find \(k\) such that \(-\frac{41\pi}{6}+ 2k\pi>0\). We know that \(2\pi=\frac{12\pi}{6}\), so \(-\frac{41\pi}{6}+ \frac{12k\pi}{6}>0\), or \(12k\pi>41\pi\), \(k > \frac{41}{12}\approx3.42\). When \(k = 4\), \(-\frac{41\pi}{6}+8\pi=-\frac{41\pi}{6}+\frac{48\pi}{6}=\frac{7\pi}{6}\).

Step2: Determine the quadrant and find reference - angle

The angle \(\theta=\frac{7\pi}{6}\) is in the third - quadrant. For an angle \(\theta\) in the third - quadrant, the reference angle \(\theta_{r}=\theta-\pi\). So \(\theta_{r}=\frac{7\pi}{6}-\pi=\frac{7\pi - 6\pi}{6}=\frac{\pi}{6}\).

Answer:

\(\frac{\pi}{6}\)