Question

$\theta = \frac{31\pi}{3}$ quadrant iv quadrant ii quadrant iii quadrant i

Explanation:

Answer:

To determine the quadrant of \( \theta = \frac{31\pi}{3} \), we first find a coterminal angle by subtracting multiples of \( 2\pi \) (since the period of the tangent and cotangent functions is \( 2\pi \), and for sine and cosine it's also \( 2\pi \)).

First, divide \( 31\pi/3 \) by \( 2\pi \) to find how many full rotations are in \( 31\pi/3 \):

\( \frac{31\pi/3}{2\pi} = \frac{31}{6} \approx 5.1667 \)

So, we can subtract \( 5 \times 2\pi = 10\pi \) (which is \( 30\pi/3 \)) from \( 31\pi/3 \):

\( \frac{31\pi}{3} - \frac{30\pi}{3} = \frac{\pi}{3} \)

Wait, that can't be right. Wait, \( 2\pi = 6\pi/3 \), so let's do it correctly. Let's find the largest integer \( n \) such that \( 2\pi n \leq 31\pi/3 \).

\( 2\pi n \leq 31\pi/3 \)

Divide both sides by \( \pi \):

\( 2n \leq 31/3 \approx 10.333 \)

So \( n = 5 \), since \( 2 \times 5 = 10 \leq 10.333 \), and \( 2 \times 6 = 12 > 10.333 \).

Now, \( 2\pi \times 5 = 10\pi = 30\pi/3 \)

Subtract that from \( 31\pi/3 \):

\( 31\pi/3 - 30\pi/3 = \pi/3 \)

Wait, \( \pi/3 \) is in Quadrant I. But that seems off. Wait, maybe I made a mistake. Let's check again.

Wait, \( 31\pi/3 = 10\pi + \pi/3 \), but \( 10\pi \) is 5 full rotations (since each rotation is \( 2\pi \)), so \( 10\pi = 5 \times 2\pi \), so the coterminal angle is \( \pi/3 \), which is in Quadrant I. But that contradicts my initial thought. Wait, maybe I miscalculated the number of rotations.

Wait, \( 2\pi = 6\pi/3 \), so let's divide \( 31\pi/3 \) by \( 6\pi/3 \) (which is \( 2\pi \)):

\( \frac{31\pi/3}{6\pi/3} = 31/6 \approx 5.1667 \), so 5 full rotations (5 times \( 6\pi/3 = 10\pi/3 \)? Wait, no, \( 2\pi = 6\pi/3 \), so each rotation is \( 6\pi/3 \). So 5 rotations would be \( 5 \times 6\pi/3 = 30\pi/3 \), which is \( 10\pi \). Then \( 31\pi/3 - 30\pi/3 = \pi/3 \), which is \( 60^\circ \), in Quadrant I. So the answer should be Quadrant I.

But wait, maybe I made a mistake. Let's check with another method. Let's convert \( 31\pi/3 \) to degrees:

\( \frac{31\pi}{3} \times \frac{180^\circ}{\pi} = 31 \times 60^\circ = 1860^\circ \)

Now, subtract multiples of \( 360^\circ \) to find the coterminal angle:

\( 1860^\circ \div 360^\circ = 5.1667 \), so 5 times \( 360^\circ = 1800^\circ \)

\( 1860^\circ - 1800^\circ = 60^\circ \), which is in Quadrant I. So the answer is Quadrant I.

So the correct option is:

Quadrant I

So the answer is the last option: Quadrant I.