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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. 89 (page 487)

Write the trigonometric expression as an algebraic expression containing uand v
$\tan (\sin^{- 1} u - \cos^{- 1} v)$

Short Answer

Expert verified

$= \frac{u v - \sqrt{1 - u^{2}} \sqrt{1 - v^{2}}}{v \sqrt{1 - u^{2}} + u \sqrt{1 - v^{2}}}$

Step by step solution

01

Step 1. The given information

trigonometric equation $\tan (\sin^{- 1} u - \cos^{- 1} v)$

02

Step 2. Putting sin-1u=xu=sinxcos-1v=yv=cosy

We know that,

$\cos x = \sqrt{1 - \sin^{2} x}$

now,

$\cos x = \sqrt{1 - u^{2}}$

Also, $\tan x = \frac{\sin x}{\cos x}$

$= \frac{u}{\sqrt{1 - u^{2}}}$

03

Step 3. Similarly

$\sin y = \sqrt{1 - v^{2}} \tan y = \frac{\sqrt{1 - v^{2}}}{v}$

04

Step 4. Using the formula tanx-y=tanx-tany1+tanxtany

Solving we get,

$= \frac{\frac{u}{\sqrt{1 - u^{2}}} - \frac{\sqrt{1 - v^{2}}}{v}}{1 + \frac{u}{\sqrt{1 - u^{2}}} \frac{\sqrt{1 - v^{2}}}{v}} = \frac{\frac{u v - \sqrt{1 - v^{2}} \sqrt{1 - u^{2}}}{v \sqrt{1 - u^{2}}}}{\frac{v \sqrt{1 - u^{2}} + u \sqrt{1 - v^{2}}}{v \sqrt{1 - u^{2}}}}$

$\frac{u v - \sqrt{(1 - u^{2}) (1 - v^{2})}}{v \sqrt{1 - u^{2}} + u \sqrt{1 - v^{2}}}$

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