Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 2 - Acute Angles and Right Triangles - Section 2.2 Trigonometric Functions of Non-Acute Angles - 2.2 Exercises - Page 62: 38


$\sin$ 1500$^{\circ}$ = $\frac{\sqrt3}{2}$

Work Step by Step

$\sin$ 1500$^{\circ}$ First, lets find the coterminal angle. 1500$^{\circ}$ - 4(360$^{\circ}$) = 60$^{\circ}$ Next, we find the reference angle. $\theta$$^{1}$ = 60$^{\circ}$ Since the coterminal angle is in Quadrant I, $\sin$ is positive. Therefore: $\sin$ 60$^{\circ}$ = $\frac{\sqrt3}{2}$
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