Trigonometry (11th Edition) Clone

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

Chapter 5 - Review Exercises - Page 249: 63


$\frac{sin^2~x-cos^2~x}{csc~x} = 2~sin^3~x-sin~x$

Work Step by Step

$\frac{sin^2~x-cos^2~x}{csc~x} = \frac{sin^2~x-cos^2~x}{\frac{1}{sin~x}}$ $\frac{sin^2~x-cos^2~x}{csc~x} = sin^3~x-sin~x~cos^2~x$ $\frac{sin^2~x-cos^2~x}{csc~x} = sin^3~x-sin~x~(1-sin^2~x)$ $\frac{sin^2~x-cos^2~x}{csc~x} = sin^3~x-sin~x~+sin^3~x$ $\frac{sin^2~x-cos^2~x}{csc~x} = 2~sin^3~x-sin~x$
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.