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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 458)

In Problems 67鈥�78,
$f (x) = \sin x, - \frac{蟺}{2} 鈮� x 鈮� \frac{蟺}{2}, g (x) = \cos x, 0 鈮� x 鈮� 蟺 and h (x) = tanx, - \frac{蟺}{2} < x < \frac{蟺}{2}$
Find the exact value of each composite function.
$h (g^{- 1} (- \frac{1}{4}))$

Short Answer

Expert verified

The required exact value of given expression is $- \sqrt{15}$ .

Step by step solution

01

Step 1. Given information.

The composite function $h (g^{- 1} (- \frac{1}{4}))$ .

02

Step 2. Here, let g-1-14=x

So,

$g (x) = - \frac{1}{4} \cos x = - \frac{1}{4} S i n c e, \cos x < 0, s o \frac{蟺}{2} 鈮� x 鈮� 蟺 Now, sinx = \sqrt{1 - \cos^{2} x} = \sqrt{1 - {(- \frac{1}{4})}^{2}} = \sqrt{1 - \frac{1}{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}$

03

Step 3. Now, find hg-1-14.

h (g^{- 1} (- \frac{1}{4})) = h (x) = \tan x = \frac{\sin x}{\cos x} = \frac{\sqrt{15} / 4}{- 1 / 4} = - \sqrt{15}

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Most popular questions from this chapter

Rewrite over a common denominator:

$\frac{1}{1 - \cos (v)} + \frac{1}{1 + \cos (v)}$

Establish each identity.

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

True or False

$\tan (2 胃) = \frac{2 \tan 胃}{1 - \tan^{2} 胃}$

True or False: $\tan (胃) \cos (胃) = \sin (胃)$ for any $胃鈮� \frac{(2 k + 1) 蟺}{2}$

solve each equation on the interval $0 鈮� 胃 < 2 蟺$

$\sin 胃 - \sqrt{3} \cos 胃 = 1$

See all solutions

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