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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. 77 (page 486)

Find the exact value of each expression.
$\cos (\tan^{- 1} \frac{4}{3} + \cos^{- 1} \frac{5}{13})$

Short Answer

Expert verified

The value of the expression is $- \frac{33}{65}$

Step by step solution

01

Step 1. Given information

The given expression is $\cos (\tan^{- 1} \frac{4}{3} + \cos^{- 1} \frac{5}{13})$

02

Step 2. Solve the expression

Let $伪 = \tan^{- 1} \frac{4}{3}$ and $尾 = \cos^{- 1} \frac{5}{13}$

So,

localid="1646557697119" role="math" $\tan 伪 = \frac{4}{3}, - \frac{蟺}{2} 鈮� 伪鈮� \frac{蟺}{2}$ and localid="1646557725031" $\cos 尾 = \frac{5}{13}, 0 鈮� 尾鈮� 鈮� 蟺$

$\sin 伪 = \frac{4}{5} \cos 伪 = \frac{3}{5}$

$\sin 尾 = \sqrt{1 - {(\frac{5}{13})}^{2}} \sin 尾 = \frac{12}{13}$

Now,

localid="1646557850672" role="math" $\cos (伪 + 尾) = \cos 伪路 \cos 尾 - \sin 伪路 \sin 尾 = \frac{3}{5} 路 \frac{5}{13} - \frac{4}{5} 路 \frac{12}{13} = - \frac{33}{65}$

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Most popular questions from this chapter

In Problems 9鈥�36, find the exact value of each expression.

$\tan [\sin^{- 1} (- \frac{1}{2})]$

$\cos (- 胃) - \cos (胃)$

If $z = \tan \frac{伪}{2}$ , show that $\sin 伪 = \frac{2 z}{1 + z^{2}}$ .

Use the fact that

$\cos \frac{蟺}{12} = \frac{1}{4} (\sqrt{6} + \sqrt{2})$

to find $\sin \frac{蟺}{24}, \cos \frac{蟺}{24}$ .

if $\sin 伪 = - \frac{4}{5}$ , $蟺 < 伪 < \frac{3 蟺}{2}$ , then $\cos 伪 =$ _____

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