Trigonometry (11th Edition) Clone

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

Chapter 6 - Review Exercises - Page 291: 30



Work Step by Step

RECALL: (1) $\theta = \arctan{x} \longrightarrow \tan{\theta}=x$ (2) Arctan is defined only ins $(-\frac{\pi}{2}, \frac{\pi}{2})$, which means the angle is either in Quadrant I or IV. (3) $\tan{\theta} = \frac{y}{x}$ (4) $\cos{\theta} = \frac{x}{r}$, where $r=\sqrt{x^2+y^2}$ Let $\theta = \arctan{3}$ This means that $\tan{\theta} = 3$. Since tangent is positive, then the angle is in Quadrant I. With $\tan{\theta} = 3=\frac{3}{1}=\frac{y}{x}$, we can set $y=3$ and $x=1$. Solve for $r$ using the formula $r=\sqrt{x^2+y^2}$ to obtain: $r=\sqrt{1^2+3^2}=\sqrt{1+9} =\sqrt{10}$ Thus, $\cos{(\arctan{3})} \\=\cos{\theta} \\= \frac{x}{r} \\= \frac{1}{\sqrt{10}} \\= \frac{1}{\sqrt{10}} \cdot \frac{\sqrt{10}}{\sqrt{10}} \\=\frac{\sqrt{10}}{10}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.