Trigonometry (11th Edition) Clone

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

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 266: 99


$sin~(2~sec^{-1}~\frac{u}{2}) = \frac{4\sqrt{u^2-4}}{u^2}$

Work Step by Step

Let $~~\theta = sec^{-1}~\frac{u}{2}$ Then $~~sec~\theta = \frac{u}{2}$ $cos~\theta = \frac{1}{sec~\theta} = \frac{2}{u}$ We can find an expression for $sin~\theta$: $sin~\theta = \sqrt{1-cos^2~\theta}$ $sin~\theta = \sqrt{1-(\frac{2}{u})^2}$ $sin~\theta = \sqrt{\frac{u^2-4}{u^2}}$ $sin~\theta = \frac{\sqrt{u^2-4}}{u}$ We can find $sin~(2\theta)$: $sin~(2\theta) = 2~sin~\theta~cos~\theta$ $sin~(2\theta) = (2)(\frac{\sqrt{u^2-4}}{u})(\frac{2}{u})$ $sin~(2\theta) = \frac{4\sqrt{u^2-4}}{u^2}$ Therefore, $~~sin~(2~sec^{-1}~\frac{u}{2}) = \frac{4\sqrt{u^2-4}}{u^2}$
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.