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: 100


$cos~(2~tan^{-1}~\frac{3}{u}) = \frac{u^2-9}{u^2+9}$

Work Step by Step

Let $~~\theta = tan^{-1}~\frac{3}{u}$ Then $~~tan~\theta = \frac{3}{u}$ We can find an expression for $sin~\theta$: $sin~\theta = \frac{3}{\sqrt{u^2+3^2}}$ $sin~\theta = \frac{3}{\sqrt{u^2+9}}$ We can find $cos~(2\theta)$: $cos~(2\theta) = 1-2~sin^2~\theta$ $cos~(2\theta) = 1-(2)(\frac{3}{\sqrt{u^2+9}})^2$ $cos~(2\theta) = 1-\frac{18}{u^2+9}$ $cos~(2\theta) = \frac{u^2-9}{u^2+9}$ Therefore, $~~cos~(2~tan^{-1}~\frac{3}{u}) = \frac{u^2-9}{u^2+9}$
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