There are four positive integers $a$, $b$, $c$, and $d$ such that \[ 4\cos(x)\cos(2x)\cos(4x) = \cos(ax) + \cos(bx) + \cos(cx) + \cos(dx) \]for all values of $x$. Answer with $a, b, c, d$ in any order, separated by commas.

[tex]4\cos{(x)}\cos{(2x)}\cos{(4x)}=2\cos{(x)}(\cos{(4x+2x)}+\cos{(4x-2x)})\\\\=\cos{(6x+x)}+\cos{(6x-x)}+\cos{(2x+x)}+\cos{(2x-x)}\\\\=\cos{(x)}+\cos{(3x)}+\cos{(5x)}+\cos{(7x)}[/tex]

So, {a, b, c, d} = {1, 3, 5, 7}.

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LammettHash LammettHash · Answer 2 · 2019-10-07T21:59:16+02:00

Overkill method: consider the expansion of [tex]e^{7ix}[/tex]:

[tex]e^{7ix}=e^{ix}e^{2ix}e^{4ix}[/tex]

[tex]\implies\cos7x+i\sin7x=(\cos x+i\sin x)(\cos2x+i\sin2x)(\cos4x+i\sin4x)[/tex]

Expand the right hand side and compare the real parts:

[tex]\cos7x=\cos x\cos2x\cos4x-\cos x\sin2x\sin4x-\sin x\cos2x\sin4x-\sin x\sin2x\cos4x[/tex]

[tex]\implies\cos x\cos2x\cos4x=\cos7x+\cos x\sin2x\sin4x+\sin x\cos2x\sin4x+\sin x\sin2x\cos4x[/tex]

Rewrite the sines in terms of cosines by using the angle sum identity.

[tex]\sin\alpha\sin\beta=\dfrac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}2[/tex]

[tex]\implies\cos x\cos2x\cos4x=\cos7x+\dfrac{\cos x(\cos2x-\cos6x)}2+\dfrac{\cos2x(\cos3x-\cos5x)}2+\dfrac{\cos4x(\cos x-\cos3x)}2[/tex]

[tex]\implies2\cos x\cos2x\cos4x=2\cos7x+\cos x\cos2x-\cos x\cos6x+\cos2x\cos3x-\cos2x\cos5x+\cos4x\cos x-\cos4x\cos3x[/tex]

Then using the other variant of the same identity,

[tex]\cos\alpha\cos\beta=\dfrac{\cos(\alpha-\beta)+\cos(\alpha+\beta)}2[/tex]

[tex]\implies2\cos x\cos2x\cos4x=2\cos7x+\dfrac{\cos x+\cos3x}2-\dfrac{\cos5x+\cos7x}2+\dfrac{\cos x+\cos5x}2-\dfrac{\cos3x+\cos7x}2+\dfrac{\cos3x+\cos5x}2-\dfrac{\cos x+\cos7x}2[/tex]

[tex]\implies4\cos x\cos2x\cos4x=4\cos7x+\cos x+\cos3x-\cos5x-\cos7x+\cos x+\cos5x-\cos3x-\cos7x+\cos3x+\cos5x-\cos x-\cos7x[/tex]

[tex]\implies4\cos x\cos2x\cos4x=\cos7x+\cos x+\cos3x+\cos5x[/tex]

so that [tex](a,b,c,d)=(7,1,3,5)[/tex].