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The Fourier series expansion for $f(x) = \sin 5x \sin x$ is $\dfrac{\cos 4x - \cos 6x}{2}$? This makes sense as $f(x) = \sin 5x \sin x$ is made up of the product of two odd functions which equals an even function and hence why there are are no sine functions in the answer.

Cheers

Amzoti
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xiA
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1 Answers1

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Note that for $x,y \in \mathbb R$, we have by the addition formulae for the cosine: $$ \cos(x \pm y) = \cos x \cos y \mp \sin x \sin y $$ Subtracting, we get $$ \cos(x+y) - \cos(x-y) = -2 \sin x \sin y $$ Now let $y = 5x$, to obtain $$ \cos(6x) - \cos(4x) =- 2 \sin x \sin 5x $$

martini
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  • Does that mean that the fourier expansion is incorrect or just my reasoning? – xiA May 31 '13 at 05:23
  • I thought and interpreted the title that you asked for a reasoning using trigonometric identities. You are correct, I have a typo, which I will correct. – martini May 31 '13 at 05:25
  • Oh yeah I see, sorry about that. Thanks – xiA May 31 '13 at 05:27