Question

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.

Answer 1

Overkill method: consider the expansion of
`e^(7ix)`:

`e^(7ix)=e^(ix)e^(2ix)e^(4ix)`

`\implies\cos7x+i\sin7x=(\cos x+i\sin x)(\cos2x+i\sin2x)(\cos4x+i\sin4x)`

Expand the right hand side and compare the real parts:

`\cos7x=\cos x\cos2x\cos4x-\cos x\sin2x\sin4x-\sin x\cos2x\sin4x-\sin x\sin2x\cos4x`

`\implies\cos x\cos2x\cos4x=\cos7x+\cos x\sin2x\sin4x+\sin x\cos2x\sin4x+\sin x\sin2x\cos4x`

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

`\sin\alpha\sin\beta=\frac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}2`

`\implies\cos x\cos2x\cos4x=\cos7x+\frac{\cos x(\cos2x-\cos6x)}2+\frac{\cos2x(\cos3x-\cos5x)}2+\frac{\cos4x(\cos x-\cos3x)}2`

`\implies2\cos x\cos2x\cos4x=2\cos7x+\cos x\cos2x-\cos x\cos6x+\cos2x\cos3x-\cos2x\cos5x+\cos4x\cos x-\cos4x\cos3x`

Then using the other variant of the same identity,

`\cos\alpha\cos\beta=\frac{\cos(\alpha-\beta)+\cos(\alpha+\beta)}2`

`\implies2\cos x\cos2x\cos4x=2\cos7x+\frac{\cos x+\cos3x}2-\frac{\cos5x+\cos7x}2+\frac{\cos x+\cos5x}2-\frac{\cos3x+\cos7x}2+\frac{\cos3x+\cos5x}2-\frac{\cos x+\cos7x}2`

`\implies4\cos x\cos2x\cos4x=4\cos7x+\cos x+\cos3x-\cos5x-\cos7x+\cos x+\cos5x-\cos3x-\cos7x+\cos3x+\cos5x-\cos x-\cos7x`

`\implies4\cos x\cos2x\cos4x=\cos7x+\cos x+\cos3x+\cos5x`

so that
`(a,b,c,d)=(7,1,3,5)`.

Answer 2

1, 3, 5, 7

Explanation:

You can use the identity ...

`2cos((a))cos((b))=cos((a+b))+cos((a-b))`

Then ...

`4cos((x))cos((2x))cos((4x))=2cos((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))`

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