This page is designed to help students, parents, and teachers master the concept of coterminal angles through clear explanations and step-by-step solutions. Understanding how to find positive and negative coterminal angles in both degrees and radians is crucial for advanced trigonometry and calculus.
Key topics covered on this page include:
Angles are called coterminal if they share the same terminal side when drawn in standard position. For example, angles \(\alpha = 30^\circ\) and \(\beta = -330^\circ\) are coterminal.
Coterminal angles \(A_c\) can be found by adding or subtracting integer multiples of a full rotation:
where \(k\) is any positive or negative integer (\(k \in \mathbb{Z}\)).
Problem: Find a positive and a negative coterminal angle for \(A = -200^\circ\).
Step 1 (Positive): Add \(360^\circ\) to get a positive value: \[A_c = -200^\circ + 360^\circ = 160^\circ\]
Step 2 (Negative): Subtract \(360^\circ\) to find a negative coterminal angle: \[A_c = -200^\circ - 360^\circ = -560^\circ\]
Answer: One positive coterminal angle is \(160^\circ\) and one negative is \(-560^\circ\).
Problem: Find a coterminal angle \(A_c\) for \(A = -\dfrac{17\pi}{3}\) such that \(0 \le A_c < 2\pi\).
Step 1: Determine how many full circles (\(2\pi\)) can be added. Notice that \(-\dfrac{17\pi}{3} = -5\dfrac{2\pi}{3}\).
Step 2: Add multiples of \(2\pi\) (or \(6\pi/3\)) until the angle falls between \(0\) and \(2\pi\): \[A_c = -\frac{17\pi}{3} + 3(2\pi) = -\frac{17\pi}{3} + 6\pi = -\frac{17\pi}{3} + \frac{18\pi}{3} = \frac{\pi}{3}\]
Answer: \(\dfrac{\pi}{3}\)
Problem: Find a coterminal angle \(A_c\) for \(A = \dfrac{35\pi}{4}\) such that \(0 \le A_c < 2\pi\).
Step 1: Express the fraction to see how many whole components of \(2\pi\) (or \(8\pi/4\)) can be subtracted: \[\frac{35\pi}{4} = \frac{32\pi}{4} + \frac{3\pi}{4} = 4(2\pi) + \frac{3\pi}{4}\]
Step 2: Subtract \(4 \times 2\pi\) (\(8\pi\)) to place the angle in the desired range: \[A_c = \frac{35\pi}{4} - 8\pi = \frac{35\pi}{4} - \frac{32\pi}{4} = \frac{3\pi}{4}\]
Answer: \(\dfrac{3\pi}{4}\)
a) For \(A = -700^\circ\), add two full rotations (\(2 \times 360^\circ = 720^\circ\)): \[-700^\circ + 720^\circ = 20^\circ\]
b) For \(B = 940^\circ\), subtract two full rotations (\(720^\circ\)) to bring it below \(360^\circ\): \[940^\circ - 720^\circ = 220^\circ\]
a) For \(A = -\dfrac{29\pi}{6}\), add \(5\pi\) (\(\dfrac{30\pi}{6}\)): \[-\frac{29\pi}{6} + \frac{30\pi}{6} = \frac{7\pi}{6}\]
b) For \(B = \dfrac{47\pi}{4}\), subtract \(5 \times 2\pi\) (\(\dfrac{40\pi}{4}\)): \[\frac{47\pi}{4} - \frac{40\pi}{4} = \frac{7\pi}{4}\]