Precalculus: Mathematics for Calculus - 6th Edition
Precalculus: Mathematics for Calculus - 6th Edition
6th Edition
ISBN: 9780840068071
Author: Stewart, James, Redlin, Lothar, Watson, Saleem
Publisher: Cengage Learning
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Chapter 7.3, Problem 101E

(a)

To determine

To show: that there is a polynomial P(t) of degree 4 such that cos4x=P(cosx) .

(a)

Expert Solution
Check Mark

Explanation of Solution

Rewrite cos4x as,

  cos(2x+2x)

Using formula for Cosine, cos(s+t)=coss⋅cost−sins⋅sint , get

  cos(2x+2x)=cos2x⋅cos2x−sin2x⋅sin2x=cos22x−sin22x

Using the Pythagorean identify, sin2x=1−cos2x , get

  cos22x−sin22x=cos22x−(1−cos22x)=cos22x−1+cos22x=2cos22x−1

Since, cos22x can be written as (cos2x)2 , hence

  2cos22x−1=2(cos2x)2−1

Using formula for cosine, cos2x=2cos2x−1 , get

  2cos22x−1=2(2cos2x−1)2−1

Using the identity (a−b)2=a2−2ab+b2 , with the first term of expression, get

  2cos22x−1=2((2cos2x)2(2cos2x)(1)+(1)2)−1

Simplifying the terms on parentheses, get

  2(4cos4x−4cos2x+1)−1

Multiplying the terms in parentheses with 2, get

  8cos4x−8cos2x+2−1=8cos4x−8cos2x+1

Thus, cos4x is a polynomial of degree 4.

(b)

To determine

To show: that there is a polynomial Q(t) of degree 5 such that cos5x=Q(cosx) .

(b)

Expert Solution
Check Mark

Explanation of Solution

Rewrite cos5x as,

  cos(3x+2x)

Using formula for Cosine, cos(s+t)=coss⋅cost−sins⋅sint , and substituting s=3x and t=2x , get

  cos(3x+2x)=cos3x⋅cos2x−sin3x⋅sin2x

Using a triple angle formula cos3x=4cos3x−3cosx and formula for cosine,

  cos2x=2cos2x−1 , with the first term of the expression and triple angle formula,

  sin3x=3sinx−4sin3x and formula for sin, sin2x=2sinx⋅cosx with the second term of the expression, get

  cos3x⋅cos2x−sin3x⋅sin2x=(4cos3x−3cosx)(2cos2x−1)−(3sinx−4sin3x)×2sinx⋅cosx=(8cos5x−4cos3x−6cos3x+3cosx)−(6sin2x⋅cosx−8sin4x⋅cosx)=(8cos5x−10cos3x+3cosx)−(6sin2x⋅cosx−8sin4x⋅cosx)

Using the Pythagorean identity, sin2x=1−cos2x in the first term of the second parentheses and writing

  sin4x=(sin2x)2=(1−cos2x)2

In the second term of the second parentheses, get

  (8cos5x−10cos3x+3cosx)−(6sin2x⋅cosx−8sin4x⋅cosx)=8cos5x−10cos3x+3cosx−((6(1−cos2x)⋅cosx−8(1−cos2x)⋅cosx))=8cos5x−10cos3x+3cosx−(6cosx−6cos3x−8cosx(1−2cos2x+cos4x))=8cos5x−10cos3x+3cosx−6cosx+6cos3x+8cosx−16cos3x+8cos5x

Combining the like terms, get

  16cos5x−20cos3x+5cosx .

Thus, cos5x is a polynomial of degree 5.

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Chapter 7.3, Problem 100E
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Chapter 7.3, Problem 102E

Chapter 7 Solutions

Precalculus: Mathematics for Calculus - 6th Edition

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