Can't get the period of the sum of a sine and a cosine

Question

$$ x(t) = 2\cos(5t+\pi/10) + 3\sin(5\pi t) $$

I'm supposing that the signal is periodic (because sine and cosine are periodic) but then;

\begin{align} P_{\sin} &= \tfrac{2}{5} \\ P_{\cos} &= \tfrac{2\pi}{5} \end{align}

How can I find the period of the sum?

Answer 1

I don't think the sum is periodic, because $\pi$ is irrational. If $T$ were the period, there would exist integers $n,m$ such that $5T=n 2\pi$ and $5\pi T=m2\pi$, but then $$\frac{5\pi T}{5T}=\frac{m2\pi}{n 2\pi}$$ or $$\pi=\frac{m}{n}$$ $\pi$ is irrational, so this is impossible.

See also this question and its answers

Answer 2

Claim 1: If $f$ and $g$ are periodic functions with periods $p$ and $q$, respectively, then the period of $f+g$ is $\operatorname{lcm}(p,q)$.

(Hover over the blank boxes to reveal sketches of the proofs, but try to prove them yourself first!)

Proof: Let $\ell = \operatorname{lcm}(p,q)$ with $\ell = ap = bq$. Since $f(t+p) = f(t)$ and $g(t+q) = g(t)$ for all $t$, $$(f+g)(t+\ell) = f(t+ap)+g(t+bq) = f(t) + g(t) = (f+g)(t)$$. Since $\ell$ is the least common multiple of the periods $p$ and $q$, any smaller positive value $0<\ell'<\ell$ cannot satisfy $(f+g)(t+\ell') = (f+g)(t)$. Hence, $\ell$ is the period of $f+g$.

Claim 2: The periods $\tfrac{2}{5}$ and $\tfrac{2\pi}{5}$ are incommensurable (their ratio is irrational), so they have no least common multiple.

The ratio of the two numbers is $\pi$, which is well-known to be irrational. Given any two incommensurable $p$ and $q$, suppose that they have a positive common multiple; say that $ap = bq$ with $a,b \in \Bbb{Z}$. Then, $$\frac{p}{q} = \frac{b}{a} \in \Bbb{Q},$$ contradicting the fact that $\frac{p}{q}$ is irrational.