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How to Express Cos 3x as a Polynomial?

  1. Aug 21, 2015 #1
    How to express ##\cos 3x## as a polynomial in ##\cos x##?
     
  2. jcsd
  3. Aug 21, 2015 #2

    DrClaude

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    Staff: Mentor

  4. Aug 21, 2015 #3
    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: Aug 21, 2015
  5. Aug 21, 2015 #4

    DrClaude

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    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##.
     
  6. Aug 21, 2015 #5

    HallsofIvy

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    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: Aug 21, 2015
  7. Aug 21, 2015 #6

    kith

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    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.
     
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