Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 5 - Trigonometric Functions - 5.3 Trigonometric Functions Values and Angle Measures - 5.3 Exercises - Page 533: 80

Answer

$\frac{\sqrt 3}{2}$, $\frac{1}{2}$, $\sqrt 3$, $\frac{\sqrt 3}{3}$, $2$, and $\frac{2\sqrt 3}{3}$.

Work Step by Step

Given $\theta=1500^\circ=4\times360^\circ+60^\circ$ (quadrant I), we have $sin\theta =sin60^\circ=\frac{\sqrt 3}{2}$, $cos\theta =cos60^\circ=\frac{1}{2}$, $tan\theta =tan60^\circ=\sqrt 3$, $cot\theta = \frac{1}{tan\theta}=\frac{\sqrt 3}{3}$, $sec\theta = \frac{1}{cos\theta}=2$, and $csc\theta = \frac{1}{sin\theta}=\frac{2\sqrt 3}{3}$.
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