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How to express ##\cos 3x## as a polynomial in ##\cos x##?

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- #1

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How to express ##\cos 3x## as a polynomial in ##\cos x##?

- #2

DrClaude

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Look up http://www.wolframalpha.com/input/?i=Cos[3x] under "Alternate forms"

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

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

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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##.##= \cos x (\cos^2 x - \sin^2 x) - \sin x (2 \sin x \cos x)##

- #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.

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.

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- #6

kith

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[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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