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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. 68 (page 496)

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

Short Answer

Expert verified

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

Step by step solution

01

Step 1 Given identity is,

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

The formulae and the properties used in this problem are,

role="math" localid="1646558096472" $\ln a - \ln b = \ln (\frac{a}{b}) \cos (2 胃) + 1 = 2 \cos^{2} 胃 \ln a^{b} = b \ln a$

02

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

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

Hence, proved.

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Most popular questions from this chapter

True or False

$\tan (2 胃) = \frac{2 \tan 胃}{1 - \tan^{2} 胃}$

Show that $c o t^{- 1} e^{v} = \tan^{- 1} e^{- v}$

Rewrite over a common denominator:

$\frac{1}{1 - \cos (v)} + \frac{1}{1 + \cos (v)}$

True or False : In establishing an identity, it is often easiest to just multiply both sides by a well-chosen nonzero expression involving the variable.

Establish each identity.

$\frac{\sin^{3} 胃 + \cos^{3} 胃}{\sin 胃 + \cos 胃} = 1 - \sin 胃 \cos 胃$

See all solutions

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