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

Answer:

  1, 3, 5, 7

Step-by-step explanation:

You can use the identity ...

  $2\cos{(a)}\cos{(b)}=\cos{(a+b)}+\cos{(a-b)}$

Then ...

$4\cos{(x)}\cos{(2x)}\cos{(4x)}=2\cos{(x)}(\cos{(4x+2x)}+\cos{(4x-2x)})\\\\=\cos{(6x+x)}+\cos{(6x-x)}+\cos{(2x+x)}+\cos{(2x-x)}\\\\=\cos{(x)}+\cos{(3x)}+\cos{(5x)}+\cos{(7x)}$

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

Answer 2

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.

$\sin\alpha\sin\beta=\dfrac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}2$

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

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

Then using the other variant of the same identity,

$\cos\alpha\cos\beta=\dfrac{\cos(\alpha-\beta)+\cos(\alpha+\beta)}2$

$\implies2\cos x\cos2x\cos4x=2\cos7x+\dfrac{\cos x+\cos3x}2-\dfrac{\cos5x+\cos7x}2+\dfrac{\cos x+\cos5x}2-\dfrac{\cos3x+\cos7x}2+\dfrac{\cos3x+\cos5x}2-\dfrac{\cos x+\cos7x}2$

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

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

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