Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 6 - 6.6 - Inverse Trigonometric Functions - 6.6 Exercises - Page 487: 130e


The identity is verified. $arcsin~x=arctan\frac{x}{\sqrt {1-x^2}}$

Work Step by Step

Let $y=arcsin~x~~$ Then: $x=sin~y~~$ $sin^2y+cos^2y=1$ $cos^2y=1-sin^2y=1-x^2$ $\frac{1}{cos^2y}=\frac{1}{1-x^2}$ $sec^2~y=\frac{1}{1-x^2}$ $sec^2~y-1=\frac{1}{1-x^2}-1=\frac{1}{1-x^2}-\frac{1-x^2}{1-x^2}=\frac{x^2}{1-x^2}$ $tan^2y=\frac{x^2}{1-x^2}$ $tan~y=\frac{x}{\sqrt {1-x^2}}$ $y=arctan\frac{x}{\sqrt {1-x^2}}=arcsin~x$
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.