Can You Find... Trigonometry Edition

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.