Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Review Exercises - Page 646: 113


$\dfrac{2}{\sqrt {4+x^2}}$

Work Step by Step

The trigonometric ratios are as follows: $\sin \theta= \dfrac{opposite}{hypotenuse}$ ; $\cos \theta= \dfrac{Adjacent}{hypotenuse}$ and $\tan \theta= \dfrac{Opposite}{Adjacent}$ Suppose $ \theta =\tan^{-1} (\dfrac{x}{2})$ $\implies \tan \theta=\dfrac{x}{2}$ Since, $\tan \theta= \dfrac{Opposite}{Adjacent}$ Now, $\cos [\tan^{-1} (\dfrac{x}{2})] =\cos \theta= \dfrac{Adjacent}{hypotenuse}$ Using the Pythagorean Theorem, we get: $=\dfrac{2}{\sqrt {(2)^2+(x)^2}}=\dfrac{2}{\sqrt {4+x^2}}$
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