Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 2 - Acute Angles and Right Triangles - Section 2.2 Trigonometric Functions of Non-Acute Angles - 2.2 Exercises: 35

Answer

$\sin$(-1860) = $\frac{-\sqrt3}{2}$ $\cos$(-1860) = $\frac{1}{2}$ $\tan$(-1860) = -$\sqrt3$ $\cot$(-1860) = $\frac{-\sqrt3}{3}$ $\csc$(-1860) = $\frac{-2\sqrt3}{3}$ $\sec$(-1860) = 2

Work Step by Step

-1860$^{\circ}$ We must first fine the coterminal angle: -1860$^{\circ}$ + 360$^{\circ}$ = -1500$^{\circ}$ -1500$^{\circ}$ + 360$^{\circ}$ = -1140$^{\circ}$ -1140$^{\circ}$ + 360$^{\circ}$ = -780$^{\circ}$ -780$^{\circ}$ + 360$^{\circ}$ = -420$^{\circ}$ -420$^{\circ}$ + 360$^{\circ}$ = -60$^{\circ}$ -60$^{\circ}$ + 360$^{\circ}$ = 300$^{\circ}$ 150 is in Quadrant IV. Therefore all the trigonometric functions are negative with the exception of $\cos$ and $\sec$. The reference angle is: $\theta$$^{1}$ = 360$^{\circ}$ - 300$^{\circ}$ = 60$^{\circ}$ $\sin$(60) = $\frac{-\sqrt3}{2}$ $\cos$(60) = $\frac{1}{2}$ $\tan$(60) = $\frac{-\sqrt3}{1}$ = -$\sqrt3$ $\cot$(60) = $\frac{1}{-\sqrt3}$ = $\frac{-\sqrt3}{3}$ $\csc$(60) = $\frac{2}{-\sqrt3}$ = $\frac{-2\sqrt3}{3}$ $\sec$(60) = $\frac{2}{1}$ = 2
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