How to plot $\frac{\sin{|2x|}}{|\sin{x}|}$?

Question

How to plot $\frac{\sin{|2x|}}{|\sin{x}|}$?
Question is from high school book, so please avoid calculus tools.
I have to say I have no idea how to solve this problem.
Only idea is to try to plot function of her fundamental period, but how can I find it?

Answer 1

Hint: $$ \sin |2x| = 2 \sin |x| \cos |x| $$

Answer 2

The denominator is easy to plot. Take the graph of $y=\sin x$ and make all ordinates be above the $x$-axis to have a series of bumps with sharp corners at the roots of $\sin x=0,$ namely where $x=πk,$ for any $k\in \mathrm Z.$

As for the numerator, note that it is even. That is $\sin |2x|=\sin 2|x|=\sin 2|-x|.$ Thus the graph of $y=\sin 2|x|$ is symmetrical about the $y$-axis. Thus it is enough to study it only on one half-axis of the abscissae. Now since $\sin 2|x|=\sin 2x$ for $x\ge 0,$ then the graph of the numerator is the graph of $\sin 2x$ for $x\ge 0$ flipped about the $y$-axis -- that is, with a sharp point where $x=0,$ and other roots where $x=\fracπ2k.$

Finally, take the quotients of ordinates. First ignore the roots of points $x=2mπ,$ where both numerator and denominator vanish simultaneously. For other values of $x,$ first note that the quotient is even as well. Thus it suffices to study it on one half-axis. Hence for $x>0,\,x\ne 2mπ,$ we have that the sign of the ordinates depends only on the sign of the numerator, whose period is $π.$ The period of the denominator is $2π.$ Hence the period of the quotient is as well $2π.$ Hence you need only find the graph in the interval $[0,2π],$ and you would have been done. Note that the roots of this quotient will occur where $x=(1+2k)\fracπ2.$