<s>[INST] What's the most elementary way to solve this trigonometric problem? 
A bead is threaded onto a light, inextensible string of length $4m$. One end of the string is fixed to a point, $A$, on a (vertical) wall. The other end of the string is attached to a point $B$ on the wall exactly $2m$ directly below $A.$ The bead is held in place so that it is at a distance of $1m$ from the wall, such that the string is taut and the plane the string is in is perpendicular to the plane the wall is in. Find the two possible vertical components of the displacement from $B$ to the bead.
This is a question I came up with, and I thought there should be some relatively simple trigonometric methods to get to the answer. I tried lots of stuff but nothing seemed to work.
Of course, we could find the equation of the ellipse that corresponds to the locus of points the bead could be when the string is taut, and then find the two values of $y$ when $x = 1$. But I'm looking for more elementary methods involving only trigonometry and Pythagoras. This is because I want the answer to be aimed at secondary school students who know elementary trigonometry only (Pythagoras, addition angle formulae, R addition formulae etc).
Thanks in advance.
 [/INST] Here's an alternative trigonometric approach that doesn't assume knowledge of the divided triangle formulae, but at the expense of additional work algebraically.

Half the battle with trigonometry is getting started: which formulae capture the key details of your set-up? If we want to do trigonometry separately on the smaller triangles $\triangle ABC$ and $\triangle BCD$, the following points are essential:

*

*$ABD$ is a straight line, so $\angle ABC$ and $\angle CBD$ are supplementary,

*The two triangles share the length $BC = x$.

So we expect to need these angles and sides in our equations. For $\triangle ABC$ we also know $AB = 2$ and $AC = 4 - x$. A way to incorporate all three sides and  $\angle ABC$ would be the law of cosines:
\begin{align}
 \cos(\angle ABC ) &= \frac{2^2 + x^2 - (4-x)^2}{2 \cdot 2 \cdot x} \\
\implies -\cos(\angle CBD ) &= \frac{8x - 12}{4x} \\
\implies \cos(\angle CBD) &= \frac{3}{x} - 2 \tag{1} \\
\end{align}
where we used the fact $\angle ABC$ and $\angle CBD$ are supplementary, because we want to relate this finding to the other triangle.
Since $\triangle BCD$ is right-angled, we can stick to basic "SOHCAHTOA" rather than anything fancy, and as discussed above we certainly want to make use of $BC = x$ and $\angle CBD$. There are two choices which both give the solution, though in very different ways.
Take the sine of $\angle CBD$; combine with cosine by harmonic addition
Since $\sin (\angle CBD) = \frac{1}{x}$ then $(1)$ implies that
$$ \cos \theta - 3 \sin \theta = -2$$
This can be solved by the harmonic addition ("R") formula to deduce an acute angle of approximately 57.7°; using $\tan (\angle CBD) = \frac{1}{y}$ we have
$$y = \frac{1}{\tan (\angle CBD) } \approx 0.63$$
Note that the harmonic addition formula also produces an obtuse solution around 159.2°, for which we obtain $y \approx -2.63$, corresponding to the bead lying 0.63 vertically above $A$ instead of 0.63 vertically below $B$.
Take the cosine of $\angle CBD$; compare results and form a quadratic
From $\triangle ABC$ we had deduced $\cos(\angle CBD) = \frac{3}{x} - 2$.
From $\triangle BCD$ we have $\cos(\angle CBD) = \frac{y}{x}$.
Comparing results we have:
\begin{align}
\frac{y}{x} &= \frac{3}{x} - 2 \\
\implies y &= 3 - 2x
\end{align}
Filling that into our diagram we see $AD = y+2 = 5-2x$. We now have all the lengths of the right-angled triangle $\triangle ADC$ in terms of $x$, so we can form a quadratic using Pythagoras. Note we could instead write the sides in terms of $y$, but there would be fractional coefficients which make unnecessarily hard work. It's easier to find $x$ then use $y = 3-2x$ to convert back to $y$ at the end.
\begin{align}
(5 - 2x)^2 + 1^2 &= (4-x)^2 \\
\implies 3x^2 -12x + 10 &= 0\\
\implies 3(x-2)^2 &= 2 \\
\implies x = 2 \pm \sqrt {\frac 2 3} &= 2 \pm \frac{\sqrt 6 }{3}
\end{align}
Note that it's the lower of these two values which yields the positive root for $y$,
$$y = 3 - 2x = -1 + \frac{2\sqrt 6 }{3} \approx 0.63$$
while the negative root
$$y= -1 - \frac{2\sqrt 6 }{3} \approx -2.63$$
corresponds to the solution above $A$, as before.
Alternative: $\angle BAC = \angle DAC$ is common to $\triangle ABC$ and $\triangle ACD$
This works like the above, but has the slight advantage it doesn't require students to know $\cos (\pi - \theta) = - \cos (\theta)$. It makes use of the fact $ABD$ is a straight line by looking at how a small triangle lies inside the larger one. Using the law of cosines in $\triangle ABC$:
\begin{align}
 \cos(\angle BAC) &= \frac{2^2 + (4-x)^2 - x^2}{2 \cdot 2 \cdot (4-x)} \\
\implies \cos(\angle BAC) &= \frac{20 - 8x}{4(4-x)} \\
\implies \cos(\angle BAC) &= \frac{5-2x}{4-x} \tag{2}
\end{align}
Then in the right-angled triangle $\triangle ACD$, we have
\begin{align}
\cos (\angle BAC) &= \frac{y+2}{4-x} \tag{3} \\
\sin (\angle BAC) &= \frac{1}{4-x} \tag{4}
\end{align}
We can use $(2)$ and $(3)$ together to write $y$ in terms of $x$.
\begin{align}
\frac{5-2x}{4-x} &= \frac{y+2}{4-x} \\
\implies 5-2x &= y + 2 \\
\implies y = 3 - 2x
\end{align}
We can apply Pythagoras to $\triangle ACD$ and solve for $x$ and $y$, as above.
Alternatively, we can use $(2)$ and $(4)$ together to relate the sine and cosine:
\begin{align} 
(2) \implies \cos(\angle BAC) &= \frac{5-2x}{4-x} \\
 &= \frac{8-2x-3}{4-x} \\
 &= 2 - 3 \cdot \frac{1}{4-x}\\
(4) \implies \cos(\angle BAC) &=  2 - 3 \sin (\angle BAC) \\
\implies \cos(\angle BAC) + 3 \sin (\angle BAC) &=  2 
\end{align}
Which can then be solved e.g. by harmonic addition to produce an acute angle of 20.8° and obtuse angle of 122.3°. We can then solve for $y$ e.g. by using $\tan(\angle BAC) = \frac{1}{2+y}$.
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