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2023 Paper 2 Q2
D: 1500.0 B: 1500.0
trigonometric identities double angle formula triple angle formula simultaneous equations tan substitution polynomial degree chebyshev polynomials

  1. The real numbers \(x\), \(y\) and \(z\) satisfy the equations \[y = \frac{2x}{1-x^2}\,,\qquad z = \frac{2y}{1-y^2}\,,\qquad x = \frac{2z}{1-z^2}\,.\] Let \(x = \tan\alpha\). Deduce that \(y = \tan 2\alpha\) and show that \(\tan\alpha = \tan 8\alpha\). Find all solutions of the equations, giving each value of \(x\), \(y\) and \(z\) in the form \(\tan\theta\) where \(-\frac{1}{2}\pi < \theta < \frac{1}{2}\pi\).
  2. Determine the number of real solutions of the simultaneous equations \[y = \frac{3x - x^3}{1-3x^2}\,,\qquad z = \frac{3y - y^3}{1-3y^2}\,,\qquad x = \frac{3z - z^3}{1-3z^2}\,.\]
  3. Consider the simultaneous equations \[y = 2x^2 - 1\,,\qquad z = 2y^2 - 1\,,\qquad x = 2z^2 - 1\,.\]
    1. Determine the number of real solutions of these simultaneous equations with \(|x| \leqslant 1\), \(|y| \leqslant 1\), \(|z| \leqslant 1\).
    2. By finding the degree of a single polynomial equation which is satisfied by \(x\), show that all solutions of these simultaneous equations have \(|x| \leqslant 1\), \(|y| \leqslant 1\), \(|z| \leqslant 1\).

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2023 Paper 3 Q4
D: 1500.0 B: 1500.0
binomial theorem trigonometric identities polynomial expansion chebyshev polynomials coefficient extraction combinatorics

Let \(n\) be a positive integer. The polynomial \(\mathrm{p}\) is defined by the identity \[\mathrm{p}(\cos\theta) \equiv \cos\big((2n+1)\theta\big) + 1\,.\]

  1. Show that \[\cos\big((2n+1)\theta\big) = \sum_{r=0}^{n} \binom{2n+1}{2r} \cos^{2n+1-2r}\theta\,(\cos^2\theta - 1)^r\,.\]
  2. By considering the expansion of \((1+t)^{2n+1}\) for suitable values of \(t\), show that the coefficient of \(x^{2n+1}\) in the polynomial \(\mathrm{p}(x)\) is \(2^{2n}\).
  3. Show that the coefficient of \(x^{2n-1}\) in the polynomial \(\mathrm{p}(x)\) is \(-(2n+1)2^{2n-2}\).
  4. It is given that there exists a polynomial \(\mathrm{q}\) such that \[\mathrm{p}(x) = (x+1)\,[\mathrm{q}(x)]^2\] and the coefficient of \(x^n\) in \(\mathrm{q}(x)\) is greater than \(0\). Write down the coefficient of \(x^n\) in the polynomial \(\mathrm{q}(x)\) and, for \(n \geqslant 2\), show that the coefficient of \(x^{n-2}\) in the polynomial \(\mathrm{q}(x)\) is \[2^{n-2}(1-n)\,.\]

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2016 Paper 2 Q6
D: 1600.0 B: 1484.0
differential equation first order ODE substitution composition of functions chain rule Chebyshev polynomials verification functional equation

This question concerns solutions of the differential equation \[ (1-x^2) \left(\frac{\d y}{\d x}\right)^2 + k^2 y^2 = k^2\, \tag{\(*\)} \] where \(k\) is a positive integer. For each value of \(k\), let \(y_k(x)\) be the solution of \((*)\) that satisfies \(y_k(1)=1\); you may assume that there is only one such solution for each value of \(k\).

  1. Write down the differential equation satisfied by \(y_1(x)\) and verify that \(y_1(x) = x\,\).
  2. Write down the differential equation satisfied by \(y_2(x)\) and verify that \(y_2(x) = 2x^2-1\,\).
  3. Let \(z(x) = 2\big(y_n(x)\big)^2 -1\). Show that \[ (1-x^2) \left(\frac{\d z}{\d x}\right)^2 +4n^2 z^2 = 4n^2\, \] and hence obtain an expression for \(y_{2n}(x)\) in terms of \(y_n(x)\).
  4. Let \(v(x) = y_n\big(y_m(x)\big)\,\). Show that \(v(x) = y_{mn}(x)\,\).

Solution:

  1. When \(k =1\), we have \((1-x^2)(y')^2 + y^2 = 1\). Notice that if \(y_1 = x\) we have \(y_1' = 1\) and \((1-x^2) \cdot 1 + x^2 = 1\) so \(y_1\) is a solution, and we are allowed to assume this is the only solution. And notice that \(y_1(1) = 1\).
  2. When \(k = 2\) we have \((1-x^2)(y')^2 + 4y^2 = 4\). Trying \(y_2 = 2x^2-1\) we see that \(y_2' = 4x\) and \((1-x^2)(4x)^2 + 4(2x^2-1)^2 = 16x^2-16x^4+16x^4-16x^2+4 = 4\). We can also check that \(y(1) = 2 \cdot 1^2 - 1 = 1\)
  3. Let \(z(x) = 2(y_n(x))^2-1\), then \begin{align*} && \frac{\d z}{\d x} &= 4y'_n(x)y_n(x) \\ \Rightarrow && LHS &= (1-x^2)\left ( \frac{\d z}{\d x} \right)^2 + 4n^2 z^2 \\ &&&= (1-x^2)16(y'_n(x))^2(y_n(x))^2 + 4n^2(2(y_n(x))^2-1)^2 \\ &&&= 16y_n^2(1-x^2) \left [\frac{n^2-n^2y_n^2}{(1-x^2)} \right] + 16n^2y_n^4-16n^2y_n^2+4n^2 \\ &&&= 4n^2 = RHS \end{align*} Therefore \(y_{2n}(x) = 2(y_n(x))^2-1 = y_2(y_n(x))\) (notice also that \(z(1) = 2(y_n(1))^2-1 = 2-1 = 1\)).
  4. Let \(v(x) = y_n(y_m(x))\) so \begin{align*} && (y_m')^2 &= \frac{m^2(1-y_m^2)}{1-x^2} \\ && (y_n')^2 &= \frac{n^2(1-y_n^2)}{1-x^2} \\ \\ && \frac{\d v}{\d x} &= y_n'(y_m(x)) \cdot y_m'(x) \\ && (1-x^2)(v')^2 &= (1-x^2) \cdot (y_n'(y_m(x)))^2 (y_m')^2 \\ &&&= (1-x^2) \cdot (y_n'(y_m(x)))^2 \left ( \frac{m^2(1-y_m^2)}{1-x^2} \right) \\ &&&= (y_n'(y_m(x)))^2 m^2(1-y_m^2) \\ &&&= \frac{n^2(1-(y_n(y_m(x)))^2)}{1-y_m^2}m^2(1-y_m^2) \\ &&&= n^2m^2(1-v^2) \end{align*} Therefore \(v\) satisfies our differential equation and \(v(1) = y_n(y_m(1)) = y_n(1) = 1\) so it must be our desired solution.
[Note: this is another question about Chebyshev polynomials, and we have proven that we can compose them nicely. This might be more easily proven as \(T_n(x) = \cos(n \cos^{-1} x)\) and so \(T_n(T_m(x)) = \cos (n \cos^{-1}( \cos (m \cos^{-1} x))) = \cos(nm \cos^{-1}x) = T_{nm}(x)\)]

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2008 Paper 3 Q5
D: 1700.0 B: 1499.3
proof by induction recurrence relation sequences polynomial identity Chebyshev polynomials quadratic equation geometric progression

The functions \({\rm T}_n(x)\), for \(n=0\), 1, 2, \(\ldots\,\), satisfy the recurrence relation \[ {\rm T}_{n+1}(x) -2x {\rm T}_n(x) + {\rm T}_{n-1}(x) =0\, \ \ \ \ \ \ \ (n\ge1). \tag{\(*\)} \] Show by induction that \[ \left({\rm T}_n(x)\right)^2 - {\rm T}_{n-1}(x) {\rm T}_{n+1}(x) = \f(x)\,, \] where \(\f(x) = \left({\rm T}_1(x)\right)^2 - {\rm T}_0(x){\rm T}_2(x)\,\). In the case \(\f(x)\equiv 0\), determine (with proof) an expression for \({\rm T}_n(x)\) in terms of \({\rm T}_0(x)\) (assumed to be non-zero) and \({\rm r}(x)\), where \({\rm r}(x) = {\rm T}_1(x)/ {\rm T}_0(x)\). Find the two possible expressions for \({\rm r}(x)\) in terms of \(x\). %Conjecture (without proof) the general form of the solution of \((*)\).

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1997 Paper 3 Q6
D: 1700.0 B: 1516.0
differential equation second order substitution Chebyshev polynomials recurrence relation trigonometric change of variable SHM

Suppose that \(y_n\) satisfies the equations \[(1-x^2)\frac{{\rm d}^2y_n}{{\rm d}x^2}-x\frac{{\rm d}y_n}{{\rm d}x}+n^2y_n=0,\] \[y_n(1)=1,\quad y_n(x)=(-1)^ny_n(-x).\] If \(x=\cos\theta\), show that \[\frac{{\rm d}^2y_n}{{\rm d}\theta^2}+n^2y_n=0,\] and hence obtain \(y_n\) as a function of \(\theta\). Deduce that for \(|x|\leqslant1\) \[y_0=1,\quad y_1=x,\] \[y_{n+1}-2xy_n+y_{n-1}=0.\]

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1989 Paper 1 Q8
D: 1500.0 B: 1516.0
complex numbers De Moivre trig identity cos nθ multiple angle formula roots of polynomials Chebyshev polynomials

By using de Moivre's theorem, or otherwise, show that

  1. \(\cos4\theta=8\cos^{4}\theta-8\cos^{2}\theta+1;\)
  2. \(\cos6\theta=32\cos^{6}\theta-48\cos^{4}\theta+18\cos^{2}\theta-1.\)
Hence, or otherwise, find all the real roots of the equation \[ 16x^{6}-28x^{4}+13x^{2}-1=0. \] [No credit will be given for numerical approximations.]

Solution: Given that \(e^{i \theta} = \cos \theta + i \sin \theta\) we must have that

  1. \begin{align*} \cos 4 \theta &= \textrm{Re} \l e^{i 4 \theta} \r \\ &= \textrm{Re} \l (\cos \theta + i \sin \theta)^4 \r \\ &= \cos^4 \theta - \binom{4}{2}\cos^2 \theta \sin^2 \theta +\sin^4 \theta \\ &= \cos^4 \theta - 6\cos^2 \theta (1-\cos^2 \theta) +(1-\cos^2 \theta)^2 \\ &= 8\cos^4 \theta - 8\cos^2 \theta + 1 \end{align*}
  2. Similarly, \begin{align*} \cos 6 \theta &= \textrm{Re} \l e^{i 6 \theta} \r \\ &= \textrm{Re} \l (\cos \theta + i \sin \theta)^6 \r \\ &= \cos^6 \theta -\binom{6}{2}\cos^4 \theta \sin^2 \theta +\binom{6}{4} \cos^2\theta \sin^4 \theta - \sin^6 \theta \\ &= \cos^6 \theta - 15 \cos^4 \theta (1-\cos^2 \theta) + 15\cos^2 \theta (1-\cos^2\theta)^2 - (1-\cos^2 \theta)^3\\ &= 31\cos^6 \theta-45\cos^4\theta+15\cos^2\theta-1+3\cos^2 \theta-3\cos^4 \theta+\cos^6 \theta \\ &= 32 \cos^6 \theta-48\cos^4 \theta+18\cos^2 \theta-1 \end{align*}
\begin{align*} 0 &= 16x^{6}-28x^{4}+13x^{2}-1\\ &= \frac12 (32x^6-56x^4+26x^2-1) \\ &= \frac12(32x^6-48x^4+18x^2-1-(8x^4-8x^2+1)) \end{align*} Therefore if \(x = \cos \theta\) then we are looking at solving \(\cos 6 \theta = \cos 4 \theta\). \(\cos 6 \theta - \cos 4 \theta = -2 \sin 5\theta \sin \theta = 0\). So we should be looking at \(\sin 5 \theta = 0\) and \(\sin \theta = 0\). \(\sin \theta = 0 \Rightarrow x = \cos \theta = \pm 1\) both of which are roots. The other roots will be \(\cos \frac{\pi}{5}, \cos \frac{2\pi}{5}\) etc but it's unclear this is an acceptable form. Alternatively, given our two roots, we can factorize \begin{align*} 0 &= 16x^{6}-28x^{4}+13x^{2}-1 \\ &= (x^2-1)(16x^4-12x^2+1) \end{align*} We can solve \(16y^2-12y+1=0\) to see that \(x^2 = \frac{3 \pm \sqrt{5}}{8}\) so our roots are: \(x = -1, 1, \pm \sqrt{\frac{3 + \sqrt{5}}{8}}, \pm \sqrt{\frac{3 -\sqrt{5}}{8}}\) (We might notice that \(3+\sqrt{5} =\l \frac{1+\sqrt{5}}{\sqrt{2}} \r^2\) so our final answer could be: \(x = -1, 1, \pm \frac{1+\sqrt{5}}{4}, \pm \frac{\sqrt{5}-1}{4}\))

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1989 Paper 3 Q5
D: 1700.0 B: 1516.0
hyperbolic functions polynomial binomial expansion inverse hyperbolic functions Chebyshev polynomials proof coefficient extraction odd function

Given that \(y=\cosh(n\cosh^{-1}x),\) for \(x\geqslant1,\) prove that \[ y=\frac{(x+\sqrt{x^{2}-1})^{n}+(x-\sqrt{x^{2}-1})^{n}}{2}. \] Explain why, when \(n=2k+1\) and \(k\in\mathbb{Z}^{+},\) \(y\) can also be expressed as the polynomial \[ a_{0}x+a_{1}x^{3}+a_{2}x^{5}+\cdots+a_{k}x^{2k+1}. \] Find \(a_{0},\) and show that

  1. \(a_{1}=(-1)^{k-1}2k(k+1)(2k+1)/3\);
  2. \(a_{2}=(-1)^{k}2(k-1)k(k+2)(2k+1)/15.\)
Find also the value of \({\displaystyle \sum_{r=0}^{k}a_{r}.}\)

Solution: Recall, \(\cosh^{-1} x = \ln (x + \sqrt{x^2-1})\) \begin{align*} \cosh(n \cosh^{-1} x) &= \frac12 \left ( \exp(n \cosh^{-1} x) + \exp(-n\cosh^{-1}x) \right) \\ &= \frac12 \left ((x + \sqrt{x^2-1})^n + (x + \sqrt{x^2-1})^{-n} \right) \\ &= \frac12 \left ((x + \sqrt{x^2-1})^n + (x - \sqrt{x^2-1})^{n} \right) \\ \end{align*} When \(n = 2k+1\) \begin{align*} \cosh(n \cosh^{-1} x)&= \frac12 \left ((x + \sqrt{x^2-1})^n + (x - \sqrt{x^2-1})^{n} \right) \\ &= \frac12 \left (\sum_{i=0}^{2k+1}\binom{2k+1}{i}x^{2k+1-i}\left ( (\sqrt{x^2-1}^{i} + (-\sqrt{x^2-1})^{i} \right) \right) \\ &=\sum_{i=0}^{k} \binom{2k+1}{2i}x^{2k+1-2i}(x^2-1)^i \\ &=\sum_{i=0}^{k} \binom{2k+1}{2i}x^{2(k-i)+1}(x^2-1)^i \\ \end{align*} Which is clearly a polynomial with only odd degree terms. \begin{align*} a_0 &= \frac{\d y}{\d x} \vert_{x=0} \\ &= \sum_{i=0}^k\binom{2k+1}{2i} \left ( (2(k-i)+1)x^{2(k-i)}(x^2-1)^i + 2i\cdot x^{2(k-i)+2}(x^2-1) \right) \\ &= \binom{2k+1}{2k} (-1)^{k} \\ &= (-1)^k(2k+1) \end{align*}

  1. \begin{align*} a_1 &= \binom{2k+1}{2k}\binom{k}{1}(-1)^{k-1}+\binom{2k+1}{2(k-1)}(-1)^{k-1} \\ &=(-1)^{k-1}\cdot ( (2k+1)k + \frac{(2k+1)\cdot 2k \cdot (2k-1)}{3!}) \\ &= (-1)^{k-1}(2k+1)k\frac{3 + 2k-1}{3} \\ &= (-1)^{k-1}2(2k+1)k (k+1) \end{align*}
  2. \begin{align*} a_2 &= \binom{2k+1}{2k} \binom{k}{2}(-1)^{k-2} + \binom{2k+1}{2(k-1)} \binom{k-1}{1} (-1)^{k-2}+\binom{2k+1}{2(k-2)} (-1)^{k-2} \\ &= \binom{2k+1}{1} \binom{k}{2}(-1)^{k-2} + \binom{2k+1}{3} \binom{k-1}{1} (-1)^{k-2}+\binom{2k+1}{5} (-1)^{k-2} \\ &= (-1)^{k} \left (\binom{2k+1}{1} \frac{k(k-1)}{2} + \binom{2k+1}{3}(k-1)+\binom{2k+1}{5} \right) \\ &= (-1)^{k} \left ( \frac{(2k+1)k(k-1)}{2} + \frac{(2k+1)k(2k-1)}{3} + \frac{(2k+1)k(2k-1)(k-1)(2k-3)}{5\cdot2\cdot3} \right) \\ &= (-1)^k (2k+1)k\frac{1}{30} \left ( 15(k-1) + 10(2k-1)+(2k-1)(k-1)(2k-3) \right) \end{align*}
\begin{align*} \sum_{r=0}^k a_k &= \frac12 \left ((1 + \sqrt{1^2-1})^n + (1 - \sqrt{1^2-1})^{n} \right) \\ &= 1 \end{align*}

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