Question

θ = -\frac{37π}{3}
quadrant i
quadrant iv
quadrant ii
quadrant iii

Explanation:

Answer:

To determine the quadrant of \(\theta = -\frac{37\pi}{3}\), we first find a coterminal angle by adding multiples of \(2\pi\) (since the period of the sine and cosine functions is \(2\pi\)).

1. Let's find how many times \(2\pi\) fits into \(\frac{37\pi}{3}\). We know that \(2\pi=\frac{6\pi}{3}\).

• Divide \(\frac{37\pi}{3}\) by \(\frac{6\pi}{3}\): \(\frac{37\pi/3}{6\pi/3}=\frac{37}{6}\approx 6.1667\).
• The largest integer less than \(\frac{37}{6}\) is \(6\). So we add \(6\times2\pi = 12\pi=\frac{36\pi}{3}\) to \(-\frac{37\pi}{3}\) to get a coterminal angle:
• \(\theta_{coterminal}=-\frac{37\pi}{3}+\frac{36\pi}{3}=-\frac{\pi}{3}\).
• We can also add \(2\pi\) (or \(\frac{6\pi}{3}\)) one more time to get a positive coterminal angle: \(-\frac{\pi}{3}+2\pi=-\frac{\pi}{3}+\frac{6\pi}{3}=\frac{5\pi}{3}\).

1. Now, we know the quadrant of an angle \(\alpha\) is determined as follows:

• Quadrant I: \(0 < \alpha<\frac{\pi}{2}\)
• Quadrant II: \(\frac{\pi}{2}<\alpha < \pi\)
• Quadrant III: \(\pi<\alpha<\frac{3\pi}{2}\)
• Quadrant IV: \(\frac{3\pi}{2}<\alpha < 2\pi\) (or \(0\) for positive angles, for negative angles we can think in terms of clock - wise rotation)
• The angle \(\frac{5\pi}{3}\) satisfies \(\frac{3\pi}{2}=\frac{9\pi}{6}\) and \(\frac{5\pi}{3}=\frac{10\pi}{6}\), and \(2\pi=\frac{12\pi}{6}\). So \(\frac{3\pi}{2}<\frac{5\pi}{3}<2\pi\), which means the angle \(\frac{5\pi}{3}\) (and thus \(-\frac{37\pi}{3}\) since they are coterminal) is in Quadrant IV.

So the answer is Quadrant IV.