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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.

2 Answers

3 votes

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).

User Abid Iqbal
by
6.7k points
3 votes

Answer:

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}.

User FastSolutions
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6.5k points