# How can one prove that integrating $\cos{(ix)}\cos{(jx)}$ cancels in Fourier Analysis?

This portion of the Wikipedia entry on Fourier Analysis details a formula, and later says that the terms for $j \ne k$ vanish. Could someone please provide a proof of this?

I actually would like to see more than one proof, but the main thing I'm concerned with is getting a deeper intuition of the inner workings of the integration/cancellation stated in the article. I'd like to be able to intuitively know why the cosines cancel, and would like even more to be able to apply the concept to similar but different situations.

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–  Hans Lundmark Mar 7 '12 at 10:19
Use this fact $$\cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B),$$ and this one $$\cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B).$$ Average these to get $$(\cos(A + B) + \cos(A - B))/2 = \cos(A)\cos(B).$$ Take $A = it$ and $B = jt$ and integrate over $[0,2\pi]$ and you are done.
Hint: $$\cos(ix)\cos(jx)=\frac12\left(\cos\left((i-j)x\right)+\cos\left((i+j)x\right)\right)$$