How to Express Cos 3x as a Polynomial?

  • Thread starter basty
  • Start date
  • #1
95
0
How to express ##\cos 3x## as a polynomial in ##\cos x##?
 

Answers and Replies

  • #3
95
0
Look up http://www.wolframalpha.com/input/?i=Cos[3x] under "Alternate forms"
How do I expand ##\cos 3x## into below form?

Alternate.png


I know that

##\cos 3x = \cos (x + 2x)##
##= \cos x \cos 2x - \sin x \sin 2x##

where

##\cos 2x = \cos^2 x - \sin^2 x##

and

##\sin 2x = 2 \sin x \cos x##

then

##= \cos x \cos 2x - \sin x \sin 2x##
##= \cos x (\cos^2 x - \sin^2 x) - \sin x (2 \sin x \cos x)##
 
Last edited:
  • #4
DrClaude
Mentor
7,540
3,873
##= \cos x (\cos^2 x - \sin^2 x) - \sin x (2 \sin x \cos x)##
Simplify the line above and you get the answer from WolframAlpha. You can then convert the ##\sin^2 x## to get a polynomial in ##\cos x##.
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,833
961
Two of the things you should know by heart are that [itex]cos(a+ b)= cos(a)cos(b)- sin(a)sin(b)[/itex] and that [itex]sin(a+ b)= sin(a)cos(b)+ cos(a)sinb)[/itex]

Taking a= b= x gives [itex]cos(2x)= cos^2(x)- sin^2(x)[/itex] and [itex]sin(2x)= 2sin(x)cos(x)[/itex].

Then [itex]cos(3x)= cos(2x+ x)= cos(2x)cos(x)- sin(2x)sin(x)[/itex] and apply the double angles identities to that.
 
Last edited by a moderator:
  • #6
kith
Science Advisor
1,370
462
If you are familiar with complex numbers, you can also use Euler's formula. [itex]\text{cos}(3x)[/itex] is the real part of
[itex]\text{e}^{i3x} = (\text{e}^{ix})^3 = (\text{cos}(x) + \text{i} \,\text{sin}(x))^3[/itex]. So if you multiply this out and take the real part, you get your answer without using further identities. This method is very powerful and general but you may not be familiar with complex numbers yet.
 

Related Threads on How to Express Cos 3x as a Polynomial?

Replies
2
Views
3K
  • Last Post
Replies
5
Views
5K
Replies
4
Views
700
  • Last Post
Replies
2
Views
14K
  • Last Post
Replies
1
Views
2K
Replies
1
Views
2K
Replies
1
Views
797
  • Last Post
Replies
4
Views
879
Replies
1
Views
1K
Replies
1
Views
1K
Top