Trigonometry (10th Edition)

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

Chapter 3 - Radian Measure and the Unit Circle - Section 3.3 The Unit Circle and Circular Functions - 3.3 Exercises - Page 118: 51

Answer

$\displaystyle \sin\theta=\frac{\sqrt{2}}{2}$ $\displaystyle \cos\theta=\frac{\sqrt{2}}{2}$ $\tan\theta=1$ $\cot\theta=1$ $\sec\theta=\sqrt{2}$ $\csc\theta=\sqrt{2}$

Work Step by Step

For any real number $s$ represented by a directed arc on the unit circle, $\sin s=y\quad \cos s=x \quad \displaystyle \tan s=\frac{y}{x} (x\neq 0)$ $\displaystyle \csc s=\frac{1}{y} (y\neq 0)\quad \displaystyle \sec s=\frac{1}{x} (x\neq 0) \displaystyle \quad\cot s=\frac{x}{y} (y\neq 0)$. ------------------- $\displaystyle \sin\theta=y=\frac{\sqrt{2}}{2}$ $\displaystyle \cos\theta=x=\frac{\sqrt{2}}{2}$ $\displaystyle \tan\theta=\frac{y}{x}=\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}=1$ $\displaystyle \cot\theta=\frac{x}{y}=\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}=1$ $\displaystyle \sec\theta=\frac{1}{x}=\frac{1}{\frac{\sqrt{2}}{2}}=\frac{2}{\sqrt{2}}=\frac{2}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\sqrt{2}$ $\displaystyle \csc\theta=\frac{1}{y}=\frac{1}{\frac{\sqrt{2}}{2}}=\frac{2}{\sqrt{2}}=\frac{2}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\sqrt{2}$
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