How to find the period of $\sin(3\pi\{x\}) + \tan(\pi[x])$

Question

Here...[.],{.} are greatest integer functions and fractional part functions

Answer 1

First note $\tan(\pi[x]) = 0$ since $\tan(n\pi) = 0$

$\sin(3\pi\{x\}) = \sin(3\pi(x - [x])) = \sin(3\pi x)\cos(3\pi[x]) - \underbrace{\cos(3\pi x)\sin(3\pi [x])}_{=0} = \sin(3\pi x)\cos(3\pi[x]) $

Let $f(x) = \sin(3\pi x)\cos(3\pi[x])$

Consider $0 \le x \le 1$. In this range $\cos(3\pi[x]) = 1$

$\sin(3\pi x)$ starts at $0$ going positive and crosses $0$ three times, at $x = 1/3, 2/3, 3/3$

Consider $1 \le x \le 2$. In this range $\cos(3\pi[x]) = -1$

$-\sin(3\pi x)$ starts at $0$ going positive and crosses $0$ three times, at $x = 4/3, 5/3, 6/3$

You will see the period is $1$

Answer 2

Deduce it in this manner: Note that the period of $tan$ is $π$. And the minimum increment of $x$ for which the argument changes is $1$. The argument obtained on replacing $x$ by $x+1$, gives us the same functional value. So the period of the tan function is $1$. Notice that you get the same value for $\{x\}$ after every increment of $1$ in the value of $x$. Thus, the period of the $sin$ function too is $1$. So the period of the entire function is the $LCM(1,1)$ which is $1$.