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1990 Paper 3 Q10
D: 1700.0 B: 1516.0
curve sketching trigonometric small angle approximation roots of equations intersection of curves inequality tangent line asymptotic expansion

By considering the graphs of \(y=kx\) and \(y=\sin x,\) show that the equation \(kx=\sin x,\) where \(k>0,\) may have \(0,1,2\) or \(3\) roots in the interval \((4n+1)\frac{\pi}{2} < x < (4n+5)\frac{\pi}{2},\) where \(n\) is a positive integer. For a certain given value of \(n\), the equation has exactly one root in this interval. Show that \(k\) lies in an interval which may be written \(\sin\delta < k < \dfrac{2}{(4n+1)\pi},\) where \(0 < \delta < \frac{1}{2}\pi\) and \[ \cos\delta=\left((4n+5)\frac{\pi}{2}-\delta\right)\sin\delta. \] Show that, if \(n\) is large, then \(\delta\approx\dfrac{2}{(4n+5)\pi}\) and obtain a second, improved, approximation.

Solution:

TikZ diagram
Clearly we can achieve \(0\), \(1\), and \(2\) intersections by never entering the range, entering too flat, or entering before hitting the second branch. To achieve \(3\) we can go at a flat enough slope that we hit somewhere near the top of the second branch, and since the gradient there will be \(\approx 0\), and our gradient is positive, we must intersect before that point as well, ie \(3\) intersections. Clearly we cannot intersect the second branch \(3\) times or the first branch twice, therefore there are at most \(3\) intersections. To intersect the graph only once, we need to:
  • be below \(((4n+1)\tfrac{\pi}{2}, 1)\) and
  • not touch the second gradient
The first condition means that \(k (4n+1)\tfrac{\pi}{2} < 1 \Rightarrow k < \frac{2}{(4n+1)\pi}\). For the second condition, consider a point on the curve \(\sin x\) whose tangent line goes through the origin, ie \(\frac{y - \sin t}{x - t} = \cos t \Rightarrow y = (\cos t)x - t \cos t+\sin t\) ie \(\sin t = t \cos t\). For this point \(t\) to be in the required interval, we need \((4n+5) \tfrac{\pi}{2} -t \in (0, \frac{\pi}{2})\), so let's call this value \(\delta\). Then our result is: The gradient needs to be steeper than \(\cos t = \cos \left ( (4n+5) \tfrac{\pi}{2} - \delta \right) = \sin \delta\) and \(\cos \delta =\left ( (4n+5) \tfrac{\pi}{2} - \delta \right) \sin \delta \). If \(n\) is large, then, \begin{align*} && 1 &\approx \left ( (4n+5) \tfrac{\pi}{2} - \delta \right) \delta \\ \Rightarrow && 1 &\approx (4n+5) \tfrac{\pi}{2} \delta \\ \Rightarrow && \delta &\approx \frac{2}{(4n+5)\pi} \end{align*}. To higher order: \begin{align*} && 1-\frac12 \delta^2 &\approx \left ( (4n+5) \tfrac{\pi}{2} - \delta \right) \delta \\ \Rightarrow && 1-\frac12 \delta^2 &\approx (4n+5) \tfrac{\pi}{2} \delta - \delta^2 \\ \Rightarrow && 0 &\approx 1 - (4n + 5)\tfrac{\pi}{2} \delta + \frac12 \delta^2 \\ \Rightarrow && \delta &\approx (4n+5) \tfrac{\pi}{2} - \sqrt{(4n+5)^2 \frac{\pi^2}{4} - 2} \\ &&&= \frac{2}{(4n+5) \tfrac{\pi}{2} + \sqrt{(4n+5)^2 \frac{\pi^2}{4} - 2}} \end{align*}.

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