Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 5 - Trigonometric Functions - Section 5.3 Properties of the Trigonometric Functions - 5.3 Assess Your Understanding - Page 418: 53


$\cos{\theta} =\dfrac{1}{2}$ $\sin{\theta} =- \dfrac{\sqrt{3}}{2}$ $\tan{\theta}=-\sqrt{3}$ $\csc{\theta} =-\dfrac{2 \sqrt{3} }{3}$ $\cot{\theta} = -\dfrac{\sqrt{3}}{3}$

Work Step by Step

$\cos{\theta} = \dfrac{1}{\sec{\theta}} = \dfrac{1}{2}$ Since $\sin{\theta} <0$, then $\sin{\theta} = -\sqrt{1-\cos^2 {\theta}}$. Thus, $\sin{\theta} = -\sqrt{1-\left(\dfrac{1}{2} \right)^2}\\ =-\sqrt{\dfrac{4}{4}-\dfrac{1}{4}}\\ =-\sqrt{\dfrac{3}{4}}\\ = - \dfrac{\sqrt{3}}{2}$ $\tan{\theta}= \dfrac{\sin{\theta}}{\cos{\theta}}$ $\tan{\theta}= \dfrac{- \dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} = -\sqrt{3}$ $\csc{\theta} = \dfrac{1}{\sin{\theta}}$ $\csc{\theta} = \dfrac{1}{- \dfrac{\sqrt{3}}{2}}= -\dfrac{2 \sqrt{3} }{3}$ $\cot{\theta} = \dfrac{1}{\tan{\theta}}$ $\cot{\theta} = \dfrac{1}{-\sqrt{3}} = -\dfrac{\sqrt{3}}{3}$
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