Solutions below are from Monika Pawlowska, Warsaw, Poland; Andrei Lazanu, Bucharest, Romania; Chris Tynan, St Bees School, Cumbria; Shu Cao, Oxford High School.

There are several ways to draw the graphs to achieve the given pattern. Can you produce the same set of graphs using the cosine function?

Here is Monika's method using reflections and translations of the graph of $\sin x$.

To form the pattern, you need functions $\pm \sin x+2n$, (I mean $\sin x+2n$ and $-\sin x+2n$ where $n$ is an integer).

The graph of $-\sin x$ is symmetrical to $\sin x$ with respect to the x-axis - when you change the sign, the function is reflected; when $n$ increases or decreases, the curve 'goes' 2 units upwards or downwards (it's translated). The graphs visible in the picture are for $n\in \{-4,-2,0,2,4\}$.

Chris's method uses only translations of the graph of $\sin x$.

First let's say $f(x)=\sin x$. It's obvious that this satisfies one of the lines given. Also, the transformations to $f(x)+a$ translate the graph $a$ units in the y direction (1) ( $a$ may be positive or negative).

Also, $f(x+a)$ translates the graph $-a$ units in the x direction (2).

Using (1), we can identify the equations of four more graphs, which will be:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{f_1(x)}& =\sin x+2\\ \multicolumn{1}{c}{f_2(x)}& =\sin x+4\\ \multicolumn{1}{c}{f_3(x)}& =\sin x-2\\ \multicolumn{1}{c}{f_4(x)}& =\sin x-4.\end{array}}$

We can also observe that the remaining 5 lines are just the above functions moved either $+a\pi$ or $-a\pi$ where $a$ is any odd number. So, a possible solution for the remaining 5 lines is:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{f_5(x)}& =\sin (x-\pi )\\ \multicolumn{1}{c}{f_6(x)}& =\sin (x-\pi )+2\\ \multicolumn{1}{c}{f_7(x)}& =\sin (x-\pi )+4\\ \multicolumn{1}{c}{f_8(x)}& =\sin (x-\pi )-2\\ \multicolumn{1}{c}{f_9(x)}& =\sin (x-\pi )-4.\end{array}}$

And this is one solution that satisfies the pattern.

NB. It can also be said that the pattern can be reproduced infinitely. This can be done by generalising our equations to the following:

${\displaystyle f[x]=\sin (x-a\pi )+b}$

where $a$ is 0 or 1 and $b$ is any even integer.

This is summed up by Shu Cao as follows:

Owing to the fact that the sine function is a periodic oscillating function, if we move it $2n\pi$ to the right or to the left parallel to the x-axis, ( $n$ being an integer), we will have the same graph. So we can write the sine function as $f(x)=\sin (x+2n\pi ).$ When the graph is turned upside down, it is because it has been moved $n\pi$ parallel to the x-axis, where $n$ is an odd integer. We can write it as $f(x)=\sin (x+n\pi ).$ When the graphs are shifted up or down parallel to the y-axis, the function is $f(x)+n$.

Therefore, we can summarize the equations of the family of graphs in the problem as $f(x)=\sin (x+z\pi )+2n$, where $z$ and $n$ are integers.