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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 7: Q. 67 (page 496)

In Problems $47 - 68$ , establish each identity.
$\ln |\sin 胃| = \frac{1}{2} (\ln |1 - \cos (2 胃)| - \ln 2)$

Short Answer

Expert verified

The identity $\ln |\sin 胃| = \frac{1}{2} (\ln |1 - \cos (2 胃)| - \ln 2)$ is established.

Step by step solution

01

Step 1 Given identity is,

$\ln |\sin 胃| = \frac{1}{2} (\ln |1 - \cos (2 胃)| - \ln 2)$

The properties and the formula used in this problem are,

$\ln (a) - \ln (b) = \ln (\frac{a}{b}) a \ln (b) = \ln b^{a} 1 - \cos (2 胃) = \sin^{2} 胃$

02

Step 2 Use the formula in the required places and prove that the expressions in both the sides are equal.

$\frac{1}{2} (\ln |1 - \cos (2 胃)| - \ln 2) = {(\ln |1 - \cos (2 胃)| - \ln 2)}^{\frac{1}{2}} = \ln {(\frac{|1 - \cos (2 胃)|}{2})}^{\frac{1}{2}} = \ln {|\sin^{2} 胃|}^{\frac{1}{2}} \frac{1}{2} (\ln |1 - \cos (2 胃)| - \ln 2) = \ln |\sin 胃|$

Hence, proved.

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