Can you find... trigonometry edition
What graphs can you make by transforming sine, cosine and tangent graphs?
Age
16 to 18
Challenge level
Problem
Think of the types of graphs you can obtain by a combination of stretches, reflections and translations of the graph $y=\sin x$. In this resource we refer to any of these graphs as a "sine graph".
Can you find ...
(a) ... a sine graph which touches the lines $y=3$ and $y=1?$
(b) ... a cosine graph which crosses the $x$-axis at $x=1$ and $x=-1$?
(c) ... a tangent graph which passes through the point $\big(\dfrac{\pi}{3},0\big)$ and for which the line $x=\dfrac{\pi}{2}$ is an asymptote?
You can use the free Desmos graphing calculator to help you find suitable graphs, but try to sketch some graphs first.
Could you include extra conditions in parts (a), (b) or (c) so that the graphs are unique?
This is an Underground Mathematics resource.
Underground Mathematics is hosted by Cambridge Mathematics. The project was originally funded by a grant from the UK Department for Education to provide free web-based resources that support the teaching and learning of post-16 mathematics.
Visit the site at undergroundmathematics.org to find more resources, which also offer suggestions, solutions and teacher notes to help with their use in the classroom.
Underground Mathematics is hosted by Cambridge Mathematics. The project was originally funded by a grant from the UK Department for Education to provide free web-based resources that support the teaching and learning of post-16 mathematics.
Visit the site at undergroundmathematics.org to find more resources, which also offer suggestions, solutions and teacher notes to help with their use in the classroom.
Student Solutions
Thanks to Julian from British School Manila, Nathan from Outwood Academy Newbold and Joshua from Dubai British school for submitting solutions to this problem.
(a) Julian noticed the following:
Since $y=\sin(x)$ touches the lines $y=-1$ and $y=1$, all we need to do is translate the graph upwards by 2 units.
So a possible solution is $y=\sin(x)+2$.
(b) Nathan and Joshua both noticed:
Since $y=\cos(x)$ crosses the $x$-axis at $\frac{\pi}{2}$ and $-\frac{\pi}{2}$, we can make it cross at 1 and -1 if we stretch the graph by a factor of $\frac{2}{\pi}$ in the $x$-direction.
So a possible solution is $y=\cos(\frac{\pi}{2} x)$.
(c) Julian and Nathan both sent us solutions for this part. Here is Julian's solution:
Since $y=\tan(x)$ already has $x=\frac{\pi}{2}$ as an asymptote, we only need to translate the graph downwards by $\tan(\frac{\pi}{3})=\sqrt{3}$ to make it pass through the point $(\frac{\pi}{3},0)$.
So a possible solution is $y=\tan(x)-\sqrt{3}$.
Nathan had a different idea:
The first step is to start by make the distance between the first $x$-intercept at the origin and the first asymptote be $\frac{\pi}{6}$ by stretching the graph by a scale factor of $\frac{1}{3}$ in the $x$-direction, and then translate the graph by $\frac{\pi}{3}$ to the right to move the $x$-intercept at the origin to $x=\frac{\pi}{3}$. This also moves the asymptote back to $x=\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{2}$.
This leads to another possible solution of $y=\tan(3(x-\frac{\pi}{3}))=\tan(3x-\pi)$.
Thomas from BHASVIC Sixth Form College Brighton has worked towards finding conditions that would specify unique solutions to these problems:
a) The general solution to this problem is $y=2+ \sin( ax+b )$. Fixing a point $1 < y_0 < 3$ at $x=0$ and fixing a gradient $ \frac{dy}{dx} $ here at $(0, y_0)$ is sufficient to specify a unique solution since only one point in each period can go through this point and have this gradient.
b) The general solution to this problem is $y=a \cos( \frac{(2n+1) \pi x}{2})$ for any integer $n$. For uniqueness we could fix the maximum value attained by the graph and the number of periods in between $x=-1$ and $x=1$.
c) For a unique solution, we only need to specify the gradient at, say, $x=\frac{ \pi }{3}$.
However, there is more than one way of making these solutions unique.