$cos(ax)=a\cdot cos(x)$ Which are all the values of $a$?

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?

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

Replacing with $$x=0$$ gives $$1 = a$$, so $$a=1$$ is the only value.

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 $$a=1$$.

• I think your argument is perfectly fine since Taylor series expansion is unique. Dec 29 '20 at 16:55

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 $$a=1$$.