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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. 87 (page 487)

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

Short Answer

Expert verified

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

Step by step solution

01

Step 1. Given information we have is

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

02

Step 2. putting tan-1u=xu=tanxsin-1v=yv=siny

$u = \frac{\sin u}{\cos u}$

$u \cos u = \sin u$

03

Step 3. Using the Pythagoras theoremsin2x+cos2x=1

${(u \cos x)}^{2} + \cos^{2} x = 1$

$u^{2} \cos^{2} x + \cos^{2} x = 1 \cos^{2} x (u^{2} + 1) = 1 \cos^{2} x = \frac{1}{u^{2} + 1} \cos x = \frac{1}{\sqrt{u^{2} + 1}}$

Now,

$\sin x = \frac{u}{\sqrt{u^{2} + 1}}$

04

Step 4. Now, we know thatcosy=1-sin2ycosy=1-v2using the formula ofsinx-y=sinxcosy-cosxsiny we get

On Solving

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

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Most popular questions from this chapter

If $x = 2 \tan 胃$ , express $\cos (2 胃)$ as a function ofx.

If $z = \tan \frac{伪}{2}$ , show that localid="1646416257885" $\cos 伪 = \frac{1 - z^{2}}{1 + z^{2}}$ .

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