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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. 27 (page 503)

In Problems 26 and 27, write each trigonometric expression as an algebraic expression in u.
$\tan (c s c^{- 1} u)$

Short Answer

Expert verified

The expression in terms ofu is $\frac{1}{\sqrt{u^{2} - 1}} .$

Step by step solution

01

Step 1. Given Information

The given expression is $\tan (c s c^{- 1} u)$

We have to write the trigonometric expression as an algebraic expression in u.

02

Step 2. Writing the expression in u

To write the expression, let $x = c s c^{- 1} u$

So, $c s c x = u$

Now,

$\tan x = \frac{\sin x}{\cos x} = \frac{1}{c s c x \sqrt{1 - \sin^{2} x}} [\sin^{2} x + \cos^{2} x = 1, \cos x = \sqrt{1 - \sin^{2} x}] = \frac{1}{c s c x \sqrt{1 - {(\frac{1}{c s c x})}^{2}}} = \frac{1}{c s c x \sqrt{\frac{c s c^{2} x - 1}{c s c^{2} x}}} = \frac{1}{\sqrt{c s c^{2} x - 1}} \tan (c s c^{- 1} u) = \frac{1}{\sqrt{u^{2} - 1}}$

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Most popular questions from this chapter

Establish each identity.

$s e c 胃 - \cos 胃 = \sin 胃 \tan 胃$

Establish each identity.

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

Rewrite over a common denominator:

$\frac{\sin (胃) + \cos (胃)}{\cos (胃)} + \frac{\sin (胃) + \cos (胃)}{\sin (胃)}$

Repeat Problem $107$ for $g (x) = \cos^{2} x$ .

Establish each identity.

$\frac{1 + \sin 胃}{1 - \sin 胃} = {(s e c 胃 + \tan 胃)}^{2}$

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