# If $c\in\mathbb R,$ then prove that $\sin(x)+\sin(cx)$ is periodic iff $c\in\mathbb Q.$

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..

• 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 $g(x)=f(x)+f(x+\pi/c).$
– user436658
Commented May 31, 2020 at 15:46
• 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. Commented May 31, 2020 at 16:42
• @Eric Wofsey For continuous non-constant functions, it is true.
– user436658
Commented May 31, 2020 at 17:42
• @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.
– Jam
Commented May 31, 2020 at 18:19
• 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. Commented May 31, 2020 at 18:57

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. $$c=2l\pi/(2k\pi)=l/k.$$

• 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 $a$.
– Jam
Commented Jun 1, 2020 at 1:21
• @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 $\sin$.
– user436658
Commented Jun 1, 2020 at 5:25
• 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?
– Jam
Commented Jun 1, 2020 at 12:41
• @Jam Please, reread it, again. And then, quote the part where I (allegedly) use "$f(x)+f(x+a)$ periodic, thus $f$ periodic," please.
– user436658
Commented Jun 1, 2020 at 14:16