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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. 21 (page 505)

use sum, difference, product, or half-angle formulas to find the exact value of each expression
$\cos (\sin^{- 1} \frac{2}{3} + \tan^{- 1} \frac{3}{2})$

Short Answer

Expert verified

$\cos (\sin^{- 1} \frac{2}{3} + \tan^{- 1} \frac{3}{2}) = \frac{2 \sqrt{5} - 6}{3 \sqrt{13}}$

Step by step solution

01

Step 1. Given expression

$\cos (\sin^{- 1} \frac{2}{3} + \tan^{- 1} \frac{3}{2})$

02

Step 2. Solving

let :

\sin^{- 1} \frac{2}{3} = 伪 \tan^{- 1} \frac{3}{2} = 尾

We need to find

$\cos (伪 + 尾) = ?$

$\sin^{- 1} \frac{2}{3} = 伪鈬� \sin 伪 = \frac{2}{3} \tan^{- 1} \frac{3}{2} = 尾鈬� \tan 尾 = \frac{3}{2}$

03

Step 3. Solving for cosα , cosβ , sinβ

$\cos 伪 = \sqrt{1 - \sin^{2} 伪} \cos 伪 = \sqrt{1 - {(\frac{2}{3})}^{2}} \cos 伪 = \frac{\sqrt{5}}{3}$

Using trigonometric triangle :

$\tan 尾 = \frac{3}{2} 鈬� \cos 尾 = \frac{2}{\sqrt{2^{2} + 3^{2}}} 鈬� \cos 尾 = \frac{2}{\sqrt{13}} 鈬� \sin 尾 = \frac{3}{\sqrt{13}}$

04

Step 4. Solving for cos(α+β)

$\cos (伪 + 尾) = \cos 伪 \cos 尾 - \sin 伪 \sin 尾 \cos (伪 + 尾) = \frac{2}{\sqrt{13}} 脳 \frac{\sqrt{5}}{3} - \frac{3}{\sqrt{13}} 脳 \frac{2}{3} \cos (伪 + 尾) = \frac{2 \sqrt{5} - 6}{3 \sqrt{13}}$

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