Trigonometry (11th Edition) Clone

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

Chapter 3 - Radian Measure and the Unit Circle - Section 3.1 Radian Measure - 3.1 Exercises - Page 105: 75

Answer

$-\sqrt3$

Work Step by Step

Convert the angle measure to degrees to obtain: $=\frac{5\pi}{3} \cdot \frac{180^o}{\pi} = 5(60^o)=300^0$ Thus, $\tan{(\frac{5\pi}{3})} \\= \tan{300^o}$ 300 degrees is in Quadrant IV so : (i) its reference angle is $360^o-300^o=60^o$ (ii) its tangent is negative From Section 2.1 (page 50) , we learned that: $\tan{60^o} = \sqrt3$ This means that: $\tan{300^o} = -\tan{60^o} = -\sqrt3$.
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