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We know period of $\sin x$ is $2π.$ So period of $\sin cx$ will be $\frac{2π}{|c|}.$ Therefore period of $(\sin x+\sin cx)$ is:

$\text{LCM of }\left(2π, \frac{2π}{|c|}\right)=\frac{\text{LCM of }(2π, 2π)}{\text{HCF of }(1,|c|)}.$

Now if $c\in\mathbb R\smallsetminus\mathbb Q,$ then HCF of $1$(rational) and $|c|$(irrational) is not possible. But since $(\sin x+\sin cx)$ is periodic, so $c$ must be rational.

Conversely if $c\in\mathbb Q,$ then period of $(\sin x+\sin cx)$ is:

$\text{LCM of }\left(2π, \frac{2π}{|c|}\right)=\frac{\text{LCM of }(2π,2π)}{\text{HCF of }(1, |c|)}, \text{ which is possible as }|c|\in\mathbb Q.$

So the period of $f(x)=\sin x+\sin cx$ is $2π.$ Which is correct since $f(x+2π)=f(x).$

Hence the statement follows.

This was my approach. But I, personally, don't like this approach that much. So is there any other direct or obvious prove than this? Please suggest..

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    $\begingroup$ While you can (with a bit of luck) define the LCM of two real numbers, your off-hand "period of $(sinx+sincx)$ is ..." (while true) is a non-trivial theorem you'd have to prove, first. Much simpler: if $f$ is periodic with period $L$, so is $g(x)=f(x)+f(x+\pi/c).$ $\endgroup$
    – user436658
    Commented May 31, 2020 at 15:46
  • $\begingroup$ It is simply not true in general that the period of a sum of two functions is the LCM of their periods. See math.stackexchange.com/questions/1079/… for instance. $\endgroup$ Commented May 31, 2020 at 16:42
  • $\begingroup$ @Eric Wofsey For continuous non-constant functions, it is true. $\endgroup$
    – user436658
    Commented May 31, 2020 at 17:42
  • $\begingroup$ @ProfessorVector For $\sin x + \left(-\sin x\right)$ it isn't so I think you need the additional stipulation that the sum actually has a fundamental period. $\endgroup$
    – Jam
    Commented May 31, 2020 at 18:19
  • $\begingroup$ Well now I'm confused. I don't know much about periodicity of functions and have very preliminary knowledge of it.. The formula, I used for calculating the period of $f(x),$ is the one that was taught in my school around 2 years ago. Although it was explicitly mentioned that the formula is not applicable always e.g., $|\sin x|+|\cos x|$ has period $\fracπ2$ instead of $π$. Now am in college and haven't read about periodicity of functions yet. I accidentally came across this statement and finding it interesting went to prove it but being unsatisfied with my own approach, posted it here. $\endgroup$ Commented May 31, 2020 at 18:57

1 Answer 1

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Suppose $f(x)$ is periodic with period $L$. This means $f(x+L)=f(x)$ for all real $x$. Then, $g_a(x)=f(x)+f(x+a)$ (with some constant $a$) is periodic with period $L$, too: $g_a(x+L)=f(x+L)+f(x+a+L)=f(x)+f(x+a)=g_a(x).$ Now, consider $f(x)=\sin x+\sin cx.$ If it is periodic with period $L$, so is $$g_\pi(x)=\sin cx + \sin(cx+c\pi)=2\sin(cx+c\pi/2)\cos(c\pi/2)$$ (since $\sin x + \sin(x+\pi)=0. $) This can be $0$, if $\cos c\pi/2=0,$ but then, $c$ is rational, we know the zeroes of $\cos$. Otherwise, we must have $L=2k\pi/c,$ we know the period of $\sin.$ Also, $g_{\pi/c}(x)$ must have period $L,$ and that's $$\sin x + \sin(x+\pi/c)=2\sin(x+\pi/(2c))\cos(\pi/(2c)).$$ Same conclusion: $\cos(\pi/(2c))=0$ (i.e. $c$ rational), or $L=2l\pi$, i.e. $c=2l\pi/(2k\pi)=l/k.$

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    $\begingroup$ I think this answer relies on the inference of periodicity not just going from $f$ to $g$ but also back from $g$ to $f$, which I believe isn't necessarily true. I think we can construct a counterexample such that $f$ is aperiodic but $f(x)+f(x+a)$ is periodic using a method similar to (Overflow Q282756), where we make $f$ takes $3$ values for $x$ determined by equivalence relations dependent on $a$. $\endgroup$
    – Jam
    Commented Jun 1, 2020 at 1:21
  • $\begingroup$ @Jam You may want to reread the answer: it relies on the special form of $g_a$ for special values of $a$. You'll find it hard to locate a counterexample for the periodicity of $\sin$. $\endgroup$
    – user436658
    Commented Jun 1, 2020 at 5:25
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    $\begingroup$ I agree that $g$ is periodic and that $f$ being periodic implies that $g$ is also. And I agree with your constraints on $c$ for the periodicity of $g$. However, I don't agree that the relationship is necessarily biconditional and can be reversed; how do you justify that $f(x)+f(x+a)$ being periodic implies that $f$ is also periodic? $\endgroup$
    – Jam
    Commented Jun 1, 2020 at 12:41
  • $\begingroup$ @Jam Please, reread it, again. And then, quote the part where I (allegedly) use "$f(x)+f(x+a)$ periodic, thus $f$ periodic," please. $\endgroup$
    – user436658
    Commented Jun 1, 2020 at 14:16

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