University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.3 - Trigonometric Functions - Exercises - Page 27: 5

Answer

$\text{Answers in the following table:}$ $\begin{array}{cccccc}{\theta} & {-\pi} & {-2 \pi / 3} & {0} & {\pi / 2} & {3 \pi / 4} \\ \hline \sin \theta & {0} & {-\frac{\sqrt{3}}{2}} & {0} & {1} & {\frac{1}{\sqrt{2}}} \\ {\cos \theta} & {-1} & {-\frac{1}{2}} & {1} & {0} & {-\frac{1}{\sqrt{2}}} \\ {\tan \theta} & {0} & {\sqrt{3}} & {0} & {\mathrm{UND}} & {-1} \\ {\cot \theta} & {\mathrm{UND}} & {\frac{1}{\sqrt{3}}} & {\mathrm{UND}} & {0} & {-1} \\ {\sec \theta} & {-1} & {-2} & {1} & {\mathrm{UND}} & {-\sqrt{2}} \\ {\sec \theta} & {\mathrm{UND}} & {-\frac{2}{\sqrt{3}}} & {\mathrm{UND}} & {1} & {\sqrt{2}}\end{array}$

Work Step by Step

$\text{Answer in the following table:}$ $\begin{array}{cccccc}{\theta} & {-\pi} & {-2 \pi / 3} & {0} & {\pi / 2} & {3 \pi / 4} \\ \hline \sin \theta & {0} & {-\frac{\sqrt{3}}{2}} & {0} & {1} & {\frac{1}{\sqrt{2}}} \\ {\cos \theta} & {-1} & {-\frac{1}{2}} & {1} & {0} & {-\frac{1}{\sqrt{2}}} \\ {\tan \theta} & {0} & {\sqrt{3}} & {0} & {\mathrm{UND}} & {-1} \\ {\cot \theta} & {\mathrm{UND}} & {\frac{1}{\sqrt{3}}} & {\mathrm{UND}} & {0} & {-1} \\ {\sec \theta} & {-1} & {-2} & {1} & {\mathrm{UND}} & {-\sqrt{2}} \\ {\sec \theta} & {\mathrm{UND}} & {-\frac{2}{\sqrt{3}}} & {\mathrm{UND}} & {1} & {\sqrt{2}}\end{array}$ We can find these values without using a calculator by considering the standard unit circle values and remembering that $\sin\theta=y$ and $\cos\theta=x$ along the unit circle of radius 1.
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.