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How do I know for which values of $a$ is true the following?$\space$

$cos(ax)=a\cdot cos(x)$ for all $x\in \mathbb{R}$


It is trivial that it is true for $a=1$. But are the more values of $a$ for which it is true?

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  • $\begingroup$ There are no other values $a \in \Bbb R$ for which this holds $\endgroup$ Dec 29 '20 at 16:13
  • $\begingroup$ And how can I prove it? @BenGrossmann $\endgroup$
    – User160
    Dec 29 '20 at 16:14
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    $\begingroup$ True for all $x\in\mathbb R$? Then substitute $x=0$. $\endgroup$
    – drhab
    Dec 29 '20 at 16:14
  • $\begingroup$ True for all $x\in \mathbb{R}$ @drhab $\endgroup$
    – User160
    Dec 29 '20 at 16:15
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    $\begingroup$ Then also for $x=0$ and substituting that we find that $1=a$. $\endgroup$
    – drhab
    Dec 29 '20 at 16:16
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Replacing with $x=0$ gives $1 = a$, so $a=1$ is the only value.

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By no means a real proof, but if you series-expand teach trig term and re-combine powers in $x$, it's pretty clear:

\begin{align*}\cos\left(ax\right) &= a \cos x \\ 1 - \frac{a^2x^2}{2!} + \frac{a^4x^4}{4!} - \cdots &= a - a\frac{x^2}{2!} + a\frac{x^4}{4!} - \cdots \\ \left(1-a\right) - \frac{x^2}{2!}\left(a^2-a\right) + \frac{x^4}{4!}\left(a^4 - a\right) - \cdots&= 0\end{align*}

For the above to hold for any $x$, the terms in parentheses must each be zero. This only works with $a=1$.

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  • $\begingroup$ I think your argument is perfectly fine since Taylor series expansion is unique. $\endgroup$ Dec 29 '20 at 16:55
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If $|a|>1$, then $|a\cos(x)|=|a|>1$ for some values of $x$, but you never have $|\cos(ax)|>1$. So, it's not true then that$$(\forall x\in\Bbb R):\cos(ax)=a\cos(x).\tag1$$And, if $|a|<1$, you never have $a\cos(x)=1$, but $\cos(ax)=1$ for some value of $x$. So, gain, it's not true that you have $(1)$.

On the other hand, if $a=-1$, it's clear that you also don't have $(1)$. So, the only solution is $a=1$.

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