Let $f(x)=\cos(x)-\cos(2x)+\cos(3x)$. Because each of terms has period dividing $2\pi$, it is straight forward that the sum does too. And by graphing, one can verify that this is the smallest period. But the sum of two functions with a given period can be a smaller period, e.g., $(\cos(x) +\cos(2x)) + (-\cos(x) +\cos(2x))$ has period $\pi$, so something more involved than just knowing the periods of the summands is required to say that the period of the sum isn’t smaller. Is there a simple argument that doesn’t require graphing to show that the period is exactly $2\pi$? Ideally I would like something that I can explain to a highschooler who does not know calculus.
There is an argument using derivatives that applies to any trigonometric polynomial. Define the support of a trigonometric sum $f(\theta)=\sum a_k e^{ik\theta}$ to be $\operatorname{supp}(f)=\{k\mid a_k\neq 0\}$. Suppose that $f(\theta)$ has finite support and period $L$. Then so does its derivative, and therefore so does $inf(\theta)-f'(\theta)=\sum a_k i(n-k)e^{ik\theta}$, which will have support $\operatorname{supp}(f)\setminus\{n\}$. By repeating this, eliminating all but a single specified term, we see that $e^{ik\theta}$ has period $L$ for every $k\in \operatorname{supp}(f)$, so $L$ must be a common multiple of the period of each summand. But any common multiple of the summand periods will be a period of the sum, so the least period of a finite trigonometric sum is the least common multiple of the periods of the summands. I do not know if this fact holds for general trigonometric sums. If anybody does know, I would be interested in that as well.
Best Answer
Note that $f(x)=-3$ if and only if $x=(2n+1)\pi$ for some $n\in\mathbb{Z}$ (assuming $x\in\mathbb{R}$). So the least period must be at least $2\pi$. Combining with $\cos x,\cos 2x,\cos 3x$ all having $2\pi$ as a period then give $2\pi$ as the least period.