Problems

2022 Paper 3 Q6

D: 1500.0 B: 1500.0

small angle approximation trigonometric identities curve sketching parametric equations limits rolling circle epicycloid

Show that when $\alpha$ is small, $\cos(\theta + \alpha) - \cos\theta \approx -\alpha\sin\theta - \frac{1}{2}\alpha^2\cos\theta$. Find the limit as $\alpha \to 0$ of \[ \frac{\sin(\theta+\alpha) - \sin\theta}{\cos(\theta+\alpha) - \cos\theta} \qquad (*) \] in the case $\sin\theta \neq 0$. In the case $\sin\theta = 0$, what happens to the value of expression $(*)$ when $\alpha \to 0$?
A circle $C_1$ of radius $a$ rolls without slipping in an anti-clockwise direction on a fixed circle $C_2$ with centre at the origin $O$ and radius $(n-1)a$, where $n$ is an integer greater than $2$. The point $P$ is fixed on $C_1$. Initially the centre of $C_1$ is at $(na, 0)$ and $P$ is at $\big((n+1)a, 0\big)$.
1. Let $Q$ be the point of contact of $C_1$ and $C_2$ at any time in the rolling motion. Show that when $OQ$ makes an angle $\theta$, measured anticlockwise, with the positive $x$-axis, the $x$-coordinate of $P$ is $x(\theta) = a(n\cos\theta + \cos n\theta)$, and find the corresponding expression for the $y$-coordinate, $y(\theta)$, of $P$.
2. Find the values of $\theta$ for which the distance $OP$ is $(n-1)a$.
3. Let $\theta_0 = \dfrac{1}{n-1}\pi$. Find the limit as $\alpha \to 0$ of \[ \frac{y(\theta_0 + \alpha) - y(\theta_0)}{x(\theta_0 + \alpha) - x(\theta_0)}\,. \] Hence show that, at the point $\big(x(\theta_0),\, y(\theta_0)\big)$, the tangent to the curve traced out by $P$ is parallel to $OP$.

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2020 Paper 3 Q6

D: 1500.0 B: 1500.0

polar coordinates curve sketching area in polar coordinates trigonometric identities small angle approximation integration

Sketch the curve $y = \cos x + \sqrt{\cos 2x}$ for $-\frac{1}{4}\pi \leqslant x \leqslant \frac{1}{4}\pi$.
The equation of curve $C_1$ in polar co-ordinates is \[ r = \cos\theta + \sqrt{\cos 2\theta} \qquad -\tfrac{1}{4}\pi \leqslant \theta \leqslant \tfrac{1}{4}\pi. \] Sketch the curve $C_1$.
The equation of curve $C_2$ in polar co-ordinates is \[ r^2 - 2r\cos\theta + \sin^2\theta = 0 \qquad -\tfrac{1}{4}\pi \leqslant \theta \leqslant \tfrac{1}{4}\pi. \] Find the value of $r$ when $\theta = \pm\frac{1}{4}\pi$. Show that, when $r$ is small, $r \approx \frac{1}{2}\theta^2$. Sketch the curve $C_2$, indicating clearly the behaviour of the curve near $r=0$ and near $\theta = \pm\frac{1}{4}\pi$. Show that the area enclosed by curve $C_2$ and above the line $\theta = 0$ is $\dfrac{\pi}{2\sqrt{2}}$.

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2010 Paper 3 Q10

D: 1700.0 B: 1500.0

second order differential equations SHM elastic string Hooke's law small angle approximation bead on wire circular motion period of oscillation

A small bead $B$, of mass $m$, slides without friction on a fixed horizontal ring of radius $a$. The centre of the ring is at $O$. The bead is attached by a light elastic string to a fixed point $P$ in the plane of the ring such that $OP = b$, where $b > a$. The natural length of the elastic string is $c$, where $c < b - a$, and its modulus of elasticity is $\lambda$. Show that the equation of motion of the bead is \[ ma\ddot \phi = -\lambda\left( \frac{a\sin\phi}{c\sin\theta}-1\right)\sin(\theta+\phi) \,, \] where $\theta=\angle BPO$ and $\phi=\angle BOP$. Given that $\theta$ and $\phi$ are small, show that $a(\theta+\phi)\approx b\theta$. Hence find the period of small oscillations about the equilibrium position $\theta=\phi =0$.

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2009 Paper 2 Q2

D: 1600.0 B: 1500.0

small angle approximation stationary points curve sketching exponential function trigonometric logarithm area approximation trapezium rule

The curve $C$ has equation \[ y= a^{\sin (\pi \e^ x)}\,, \] where $a>1$.

Find the coordinates of the stationary points on $C$.
Use the approximations $\e^t \approx 1+t$ and $\sin t \approx t$ (both valid for small values of $t$) to show that \[ y\approx 1-\pi x \ln a \; \] for small values of $x$.
Sketch $C$.
By approximating $C$ by means of straight lines joining consecutive stationary points, show that the area between $C$ and the $x$-axis between the $k$th and $(k+1)$th maxima is approximately \[ \Big( \frac {a^2+1}{2a} \Big) \ln \Big ( 1+ \big( k-\tfrac34)^{-1} \Big)\,. \]

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2006 Paper 1 Q4

D: 1500.0 B: 1514.2

curve sketching inequality trigonometric differentiation regular polygon optimisation small angle approximation geometric proof

By sketching on the same axes the graphs of $y=\sin x$ and $y=x$, show that, for $x>0$:

$x>\sin x\,$;
$\dfrac {\sin x} {x} \approx 1$ for small $x$.

A regular polygon has $n$ sides, and perimeter $P$. Show that the area of the polygon is \[ \displaystyle \frac{P^2} { {4n \tan \l\dfrac{ \pi} { n} \r}} \;. \] Show by differentiation (treating $n$ as a continuous variable) that the area of the polygon increases as $n$ increases with $P$ fixed. Show also that, for large $n$, the ratio of the area of the polygon to the area of the smallest circle which can be drawn around the polygon is approximately $1$.

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2000 Paper 1 Q9

D: 1500.0 B: 1515.2

projectiles non-inertial frame accelerating frame optimisation trigonometric identity tan 2θ small angle approximation maximum range

A child is playing with a toy cannon on the floor of a long railway carriage. The carriage is moving horizontally in a northerly direction with acceleration $a$. The child points the cannon southward at an angle $\theta$ to the horizontal and fires a toy shell which leaves the cannon at speed $V$. Find, in terms of $a$ and $g$, the value of $\tan 2\theta$ for which the cannon has maximum range (in the carriage). If $a$ is small compared with $g$, show that the value of $\theta$ which gives the maximum range is approximately \[ \frac \pi 4 + \frac a {2g}, \] and show that the maximum range is approximately $\displaystyle \frac {V^2} g + \frac {V^2a}{g^2}. $

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2000 Paper 2 Q3

D: 1600.0 B: 1484.0

small angle approximation cosine rule triangle linear approximation error bounds optimisation perturbation

The lengths of the sides $BC$, $CA$, $AB$ of the triangle $ABC$ are denoted by $a$, $b$, $c$, respectively. Given that $$ b = 8+{\epsilon}_1, \, c=3+{\epsilon}_2,\, A=\tfrac{1}{3}\pi + {\epsilon}_3, $$ where ${\epsilon}_1$, ${\epsilon}_2$, and $ {\epsilon}_3$ are small, show that $a \approx 7 + {\eta}$, where ${\eta}= {\left(13 \, {{\epsilon}_1}-2\,{\epsilon}_2 + 24{\sqrt 3} \;{{\epsilon}_3}\right)}/14$. Given now that $$ {\vert {\epsilon}_1} \vert \le 2 \times 10^{-3}, \ \ \ {\vert {\epsilon}_2} \vert \le 4\cdot 9\times 10^{-2}, \ \ \ {\vert {\epsilon}_3} \vert \le \sqrt3 \times 10^{-3}, $$ find the range of possible values of ${\eta}$.

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2000 Paper 2 Q5

D: 1600.0 B: 1470.2

integration optimisation least squares approximation differentiation under integral trigonometric small angle approximation Taylor expansion integration by parts

It is required to approximate a given function $\f(x)$, over the interval $0 \le x \le 1$, by the linear function $\lambda x$, where $\lambda$ is chosen to minimise \[ \int_0^1 \big(\f(x)-\lambda x \big)^{\!2} \,\d x . \] Show that \[ \lambda = 3 \int_0^1 x\f(x)\,\d x. \] The residual error, $R$, of this approximation process is such that \[ R^2 = \int_0^1 \big(\f(x)-\lambda x \big)^{\!2}\,\d x. \] Show that \[ R^2 = \int_0^1 \big(\f(x)\big)^{\!2}\,\d x -\tfrac{1}{3} \lambda ^2. \] Given now that $\f(x)= \sin (\pi x/n)$, show that (i) for large $n$, $\lambda \approx \pi/n$ and (ii) $\lim_{n \to \infty}R = 0.$ Explain why, prior to any calculation, these results are to be expected. [You may assume that, when $\theta$ is small, $\sin \theta \approx \theta-\frac{1}{6}\theta^3$ and $\cos \theta \approx 1 - \frac{1}{2}\theta^2.$]

Solution: \begin{align*} && g(\lambda) &= \int_0^1 \big(\f(x)-\lambda x \big)^{\!2} \,\d x \\ &&&= \int_0^1 \left ( f(x)^2 -2\lambda xf(x) + \lambda^2 x^2\right) \d x \\ &&&= \frac13\lambda^2 - 2\lambda \int_0^1 x f(x) \d x + \int_0^1 f(x)^2 \d x \\ \end{align*} Differentiating (or completing the square) it is clear the minimum occurs when $\displaystyle \lambda = 3 \int_0^1 xf(x) \d x$ \begin{align*} && R^2 &= \int_0^1 (f(x) - \lambda x )^2 \d x \\ &&&= \frac13\lambda^2 - 2\lambda \int_0^1 x f(x) \d x + \int_0^1 f(x)^2 \d x \\ &&&= \frac13 \left (\lambda -3\int_0^1 xf(x) \d x \right)^2 -\frac13 \left ( 3\int_0^1 xf(x) \d x \right)^2+\int_0^1 f(x)^2 \d x \\ \end{align*} When $\lambda = 3\int_0^1 xf(x) \d x $ clearly this is the desired result. \begin{align*} && \lambda &= 3\int_0^1 xf(x) \d x \\ &&&= 3\int_0^1 x \sin(\pi x /n) \d x \\ &&&= 3 \left [-x \frac{n}{\pi} \cos (\pi x /n) \right]_0^1 + \frac{3n}{\pi} \int_0^1 \cos(\pi x /n) \d x \\ &&&= -\frac{3n}{\pi}\cos(\pi/n) + \frac{3n}{\pi} \left [ \frac{n}{\pi} \sin(\pi x /n)\right]_0^1 \\ &&&= -\frac{3n}{\pi} \cos(\pi/n) + \frac{3n^2}{\pi^2} \sin(\pi /n) \\ \text{for large }n: &&&\approx -\frac{3n}{\pi}\left ( 1 - \frac12\frac{\pi^2}{n^2} + o(1/n^4)\right) + \frac{3n^2}{\pi^2} \left (\frac{\pi}{n} - \frac16 \frac{\pi^3}{n^3} +o(1/n^5) \right) \\ &&&= \left (\frac32 -\frac12\right)\frac{\pi}{n} + o(1/n^3) \\ &&&= \frac{\pi}{n} + o(1/n^2) \end{align*} Therefore for large $n$, $\lambda \approx \frac{\pi}n$ \begin{align*} && \int_0^1 \sin^2(\pi x/n) \d x &= \frac12\int_0^1(1- \cos(2\pi x/n)) \d x\\ &&&= \frac12\left ( 1 - \frac{n}{2\pi}\left[\sin(2\pi x/n) \right]_0^1 \right) \\ &&&= \frac12 -\frac{n}{4\pi}\sin(2\pi /n) \\ \\ && R^2 &= \frac12 -\frac{n}{4\pi}\sin(2\pi /n) - \frac13 \left ( \frac{\pi}{n}+o(1/n^2)\right)^2 \\ &&&= \frac12 - \left ( \frac{1}{2} -\frac16\frac{\pi}{n}+o(1/n^3) \right) - o(1/n^2) \\ &&& = \frac16 \frac{\pi}{n} + o(1/n^2) \\ &&&\to 0 \text{ as } n \to \infty \end{align*} We should expect these results as for $n$ very large $\sin(\pi x/n) \approx \frac{\pi }{n}x$ so the best linear approximation is likely to be $\lambda = \frac{\pi}{n}$ and we should expect it to improve to the point that we cannot tell the difference, ie $R^2 \to 0$

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1999 Paper 1 Q5

D: 1500.0 B: 1516.0

small angle approximation trigonometry circle geometry satellite binomial approximation Taylor expansion arc length optimisation

For this question, you may use the following approximations, valid if $\theta $ is small: \ $\sin\theta \approx \theta$ and $\cos\theta \approx 1-\theta^2/2\,$. A satellite $X$ is directly above the point $Y$ on the Earth's surface and can just be seen (on the horizon) from another point $Z$ on the Earth's surface. The radius of the Earth is $R$ and the height of the satellite above the Earth is $h$.

Find the distance $d$ of $Z$ from $Y$ along the Earth's surface.
If the satellite is in low orbit (so that $h$ is small compared with $R$), show that $$d \approx k(Rh)^{1/2},$$ where $k$ is to be found.
If the satellite is very distant from the Earth (so that $R$ is small compared with $h$), show that $$d\approx aR+b(R^2/h),$$ where $a$ and $b$ are to be found.

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1999 Paper 2 Q8

D: 1600.0 B: 1500.0

trigonometric series summation differentiation small angle approximation proof sin sum formula asymptotic approximation product-to-sum

Prove that $$ \sum_{k=0}^n \sin k\theta = \frac { \cos \tfrac12\theta - \cos (n+ \tfrac12) \theta} {2\sin \tfrac12\theta}\;. \tag{*}$$

Deduce that, when $n$ is large, \[ \sum_{k=0}^n \sin \left(\frac{k\pi}{n}\right) \approx \frac{2n}\pi\;. \]
By differentiating $(*)$ with respect to $\theta$, or otherwise, show that, when $n$ is large, \[ \sum_{k=0}^n k \sin^2 \left(\frac{k\pi}{2n}\right) \approx \left(\frac{1}4 +\frac{1}{\pi^2} \right)n^2\;. \]

Solution: \begin{align*} && \sum_{k=0}^n \sin k\theta &= \textrm{Im} \left ( \sum_{k=0}^n e^{i k \theta}\right)\\ &&&= \textrm{Im} \left ( \frac{e^{i(n+1)\theta}-1}{e^{i \theta}-1} \right)\\ &&&= \textrm{Im} \left ( \frac{e^{i(n+\tfrac12)\theta}-e^{-i\theta/2}}{e^{i \theta/2}-e^{-i \theta/2}} \right)\\ &&&= \textrm{Im} \left ( \frac{(e^{i(n+\tfrac12)\theta}-e^{-i\theta/2})/2i}{(e^{i \theta/2}-e^{-i \theta/2})/2i} \right)\\ &&&= \textrm{Im} \left ( \frac{(e^{i(n+\tfrac12)\theta}-e^{-i\theta/2})/i}{2\sin \tfrac12 \theta} \right)\\ &&&= \frac{\cos \tfrac12 \theta - \cos(n+ \tfrac12)\theta}{2\sin \tfrac12 \theta} \end{align*}

When $n$ is large we have \begin{align*} &&\sum_{k=0}^n \sin \left(\frac{k\pi}{n}\right) &= \frac{\cos \frac{\pi}{2n} - \cos \frac{(2n + 1)\pi}{2n}}{2 \sin \frac{\pi}{2n}} \\ &&&= \frac{\cos \frac{\pi}{2n} +\cos \frac{\pi}{2n}}{2 \sin \frac{\pi}{2n}} \\ &&&\approx \frac{1- \frac{\pi^2}{4n^2}}{\frac{\pi}{2n}} \\ &&&= \frac{2n}{\pi} - \frac{\pi}{2n} \\ &&&\approx \frac{2n}{\pi} \end{align*}
$\,$ \begin{align*} && \sum_{k=0}^n \sin k\theta &= \frac { \cos \tfrac12\theta - \cos (n+ \tfrac12) \theta} {2\sin \tfrac12\theta} \\ \frac{\d }{\d \theta}: && \sum_{k=0}^n k\cos k\theta &= \frac { (-\tfrac12\sin\tfrac12\theta +(n+\tfrac12) \sin (n+ \tfrac12) \theta)2 \sin \tfrac12 \theta - (\cos \tfrac12\theta - \cos (n+ \tfrac12) \theta) \cos \tfrac12 \theta} {4\sin^2 \tfrac12\theta} \\ &&&= \frac{-\sin^2 \tfrac12 \theta+(2n+1)\sin(n+\tfrac12)\theta \sin \tfrac12 \theta-\cos^2\tfrac12 \theta+\cos(n+\frac12)\theta \cos\tfrac12 \theta}{4 \sin^2 \tfrac12 \theta} \\ &&&= \frac{(2n+1)\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos(n+\frac12)\theta \cos\tfrac12 \theta-1}{4 \sin^2 \tfrac12 \theta} \\ &&&= \frac{2n\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos n \theta-1}{4 \sin^2 \tfrac12 \theta} \\ \\ \Rightarrow && \sum_{k=0}^n k\left ( 1-2\sin^2 \left ( \frac{k\theta}{2} \right) \right) &= \frac{2n\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos n \theta-1}{4 \sin^2 \tfrac12 \theta} \\ \Rightarrow && \sum_{k=0}^n k\sin^2 \left ( \frac{k\theta}{2} \right) &= \frac{n(n+1)}{4} - \frac{2n\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos n \theta-1}{8 \sin^2 \tfrac12 \theta} \\ \theta = \frac{\pi}{n}: && \sum_{k=0}^n k\sin^2 \left ( \frac{k\pi}{2n} \right) &= \frac{n(n+1)}{4} - \frac{2n\sin(n+\tfrac12)\frac{\pi}{n}\sin \tfrac12 \frac{\pi}{n}+\cos n \frac{\pi}{n}-1}{8 \sin^2 \tfrac12 \frac{\pi}{n}} \\ &&&= \frac{n(n+1)}{4} - \frac{-2n \sin^2 \frac{\pi}{2n}-2}{8 \sin^2 \tfrac12 \frac{\pi}{n}} \\ &&&= \frac{n^2+n}{4} + \frac{n}{4} + \frac{1}{4\sin^2\frac{\pi}{2n}} \\ &&&\approx \frac{n^2}4 + \frac{n}{2}+ \frac{n^2}{\pi^2} \\ &&&= \left ( \frac14 + \frac{1}{\pi^2} \right)n^2 + \frac{n}{4} \\ &&&\approx \left ( \frac14 + \frac{1}{\pi^2} \right)n^2 \end{align*}

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1992 Paper 1 Q6

D: 1500.0 B: 1519.4

small angle approximation linear approximation trigonometry parallax error analysis differentiation distant point telescopes

Explain briefly, by means of a diagram, or otherwise, why \[ \mathrm{f}(\theta+\delta\theta)\approx\mathrm{f}(\theta)+\mathrm{f}'(\theta)\delta\theta, \] when $\delta\theta$ is small. Two powerful telescopes are placed at points $A$ and $B$ which are a distance $a$ apart. A very distant point $C$ is such that $AC$ makes an angle $\theta$ with $AB$ and $BC$ makes an angle $\theta+\phi$ with $AB$ produced. (A sketch of the arrangement is given in the diagram.) \noindent

\psset{xunit=0.8cm,yunit=0.8cm,algebraic=true,dimen=middle,dotstyle=o,dotsize=3pt 0,linewidth=0.5pt,arrowsize=3pt 2,arrowinset=0.25} \begin{pspicture*}(-4.18,-0.94)(4.4,5.22) \psline(-4,0)(4,0) \psline(-2,0)(2,5) \psline(2,5)(1,0) \rput[tl](-2.3,-0.14){$A$} \rput[tl](1.08,-0.14){$B$} \rput[tl](-1.6,0.46){$\theta$} \rput[tl](1.24,0.52){$\theta+\phi$} \rput[tl](2.14,5.1){$C$} \end{pspicture*} \par

If the perpendicular distance $h$ of $C$ from $AB$ is very large compared with $a$ show that $h$ is approximately $(a\sin^{2}\theta)/\phi$ and find the approximate value of $AC$ in terms of $a,\theta$ and $\phi.$ It is easy to show (but you are not asked to show it) that errors in measuring $\phi$ are much more important than errors in measuring $\theta.$ If we make an error of $\delta\phi$ in measuring $\phi$ (but measure $\theta$ correctly) what is the approximate error in our estimate of $AC$ and, roughly, in what proportion is it reduced by doubling the distance between $A$ and $B$?

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1992 Paper 2 Q1

D: 1600.0 B: 1500.0

limits small angle approximation squeeze theorem rational function trigonometric sinc function arctan surds rationalisation

Find the limit, as $n\rightarrow\infty,$ of each of the following. You should explain your reasoning briefly. \begin{alignat*}{4} \mathbf{(i)\ \ } & \dfrac{n}{n+1}, & \qquad & \mathbf{(ii)\ \ } & \dfrac{5n+1}{n^{2}-3n+4}, & \qquad & \mathbf{(iii)\ \ } & \dfrac{\sin n}{n},\\ \\ \mathbf{(iv)\ \ } & \dfrac{\sin(1/n)}{(1/n)}, & & \mathbf{(v)}\ \ & (\arctan n)^{-1}, & & \mathbf{(vi)\ \ } & \dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+2}-\sqrt{n}}. \end{alignat*}

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1990 Paper 1 Q8

D: 1500.0 B: 1516.0

trigonometric identity double angle formula proof by induction infinite product small angle approximation Vieta product limit

Show that \[ \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)=\frac{\sin\alpha}{4\sin\left(\dfrac{\alpha}{4}\right)}\,, \] where $\alpha\neq k\pi$, $k$ is an integer. Prove that, for such $\alpha$, \[ \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right)=\frac{\sin\alpha}{2^{n}\sin\left(\dfrac{\alpha}{2^{n}}\right)}\,, \] where $n$ is a positive integer. Deduce that \[ \alpha=\frac{\sin\alpha}{\cos\left(\dfrac{\alpha}{2}\right)\cos\left(\dfrac{\alpha}{4}\right)\cos\left(\dfrac{\alpha}{8}\right)\cdots}\,, \] and hence that \[ \frac{\pi}{2}=\frac{1}{\sqrt{\frac{1}{2}}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}}\cdots}\,. \]

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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:

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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