Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.1 - Verifying Trigonometric Identities - Exercise Set - Page 659: 29


See the explanation below.

Work Step by Step

$1-\frac{{{\sin }^{2}}x}{1+\cos x}=\cos x$ Recall Pythagorean Identity, ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ Use the above identity and solve the left side of the given expression, $1-\frac{{{\sin }^{2}}x}{1+\cos x}=\cos x$ Multiply numerator and denominator of $\frac{{{\sin }^{2}}x}{1+\cos x}$ by $1-\cos x$. $\begin{align} & 1-\frac{{{\sin }^{2}}x}{1+\cos x}=1-\frac{{{\sin }^{2}}x}{1+\cos x}\cdot \left( \frac{1-\cos x}{1-\cos x} \right) \\ & =1-\frac{{{\sin }^{2}}x\left( 1-\cos x \right)}{1-{{\cos }^{2}}x} \\ & =1-\frac{{{\sin }^{2}}x\left( 1-\cos x \right)}{{{\sin }^{2}}x} \\ & =1-\left( 1-\cos x \right) \end{align}$ Further simplify, $\begin{align} & 1-\frac{{{\sin }^{2}}x}{1+\cos x}=1-1+\cos x \\ & =\cos x \end{align}$ Therefore, $1-\frac{{{\sin }^{2}}x}{1+\cos x}=\cos x$
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