Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 5 - Section 5.5 - Inverse Trigonometric Functions and Their Graphs - 5.5 Exercises - Page 445: 48

Answer

$-\frac{\sqrt3}{2}$

Work Step by Step

First, we solve tan$^{-1}(-\sqrt3)$. The interval of the inverse of tangent is $-\frac{\pi}{2}\lt x\lt\frac{\pi}{2}$. Then, we must find what angle will give a slope of $-\sqrt3$ in that interval. The answer is $-\frac{\pi}{3}$ Now, we can solve sin$\left(-\frac{\pi}{3}\right)$ which is $-\frac{\sqrt3}{2}$
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.