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2021 Paper 3 Q7
D: 1500.0 B: 1500.0
complex numbers unit circle cotangent modulus and argument perpendicularity orthocentre geometric transformations

  1. Let \[ z = \frac{e^{i\theta} + e^{i\phi}}{e^{i\theta} - e^{i\phi}}, \] where \(\theta\) and \(\phi\) are real, and \(\theta - \phi \neq 2n\pi\) for any integer \(n\). Show that \[ z = i\cot\!\bigl(\tfrac{1}{2}(\phi - \theta)\bigr) \] and give expressions for the modulus and argument of \(z\).
  2. The distinct points \(A\) and \(B\) lie on a circle with radius \(1\) and centre \(O\). In the complex plane, \(A\) and \(B\) are represented by the complex numbers \(a\) and \(b\), and \(O\) is at the origin. The point \(X\) is represented by the complex number \(x\), where \(x = a + b\) and \(a + b \neq 0\). Show that \(OX\) is perpendicular to \(AB\). If the distinct points \(A\), \(B\) and \(C\) in the complex plane, which are represented by the complex numbers \(a\), \(b\) and \(c\), lie on a circle with radius \(1\) and centre \(O\), and \(h = a + b + c\) represents the point \(H\), then \(H\) is said to be the orthocentre of the triangle \(ABC\).
  3. The distinct points \(A\), \(B\) and \(C\) lie on a circle with radius \(1\) and centre \(O\). In the complex plane, \(A\), \(B\) and \(C\) are represented by the complex numbers \(a\), \(b\) and \(c\), and \(O\) is at the origin. Show that, if the point \(H\), represented by the complex number \(h\), is the orthocentre of the triangle \(ABC\), then either \(h = a\) or \(AH\) is perpendicular to \(BC\).
  4. The distinct points \(A\), \(B\), \(C\) and \(D\) (in that order, anticlockwise) all lie on a circle with radius \(1\) and centre \(O\). The points \(P\), \(Q\), \(R\) and \(S\) are the orthocentres of the triangles \(ABC\), \(BCD\), \(CDA\) and \(DAB\), respectively. By considering the midpoint of \(AQ\), show that there is a single transformation which maps the quadrilateral \(ABCD\) onto the quadrilateral \(QRSP\) and describe this transformation fully.

Solution:

  1. \(\,\) \begin{align*} && z &= \frac{e^{i \theta} + e^{i \phi}}{e^{i \theta} - e^{i \phi}} \\ &&&= \frac{e^{i\frac12(\theta +\phi)}(e^{i \frac12(\theta-\phi)} + e^{-i\frac12(\theta- \phi)})}{e^{i\frac12(\theta +\phi)}(e^{i \frac12(\theta-\phi)} - e^{-i\frac12(\theta- \phi)})} \\ &&&= \frac{(e^{i \frac12(\theta-\phi)} + e^{-i\frac12(\theta- \phi)})/2}{i(e^{i \frac12(\theta-\phi)} - e^{-i\frac12(\theta- \phi)})/2i} \\ &&&= \frac{\cos \frac12(\theta-\phi)}{i \sin \frac12(\theta-\phi)} \\ &&&= -i \cot \tfrac12(\theta-\phi) \end{align*} Therefore \(|z| = \cot \tfrac12(\theta-\phi)\) and \(\arg z = \frac{\pi}{2}\)
  2. Since \(a,b\) lie on the unit circle, wlog \(a = e^{i \theta}, b = e^{i\phi}\). Not that the line \(OX\) has vector \(a+b\) and \(AB\) has vector \(b-a\) and not their ratio has argument \(\frac{\pi}{2}\) and hence they are perpendicular.
  3. \(AH\) has vector \(h - a = (a+b+c) - a = b+c\) which we've already established is perpendicular to \(c-b\) which is the vector for \(BC\) (unless \(b+c = 0\) in which case \(h = a\)).
  4. \(p = a +b+c, q = b+c+d\) etc. The midpoint of \(AQ = \frac12(a+b+c+d)\) which is the same as the midpoint of \(BR\), \(CS\) and \(DP\). Therefore we could say the transformation is reflection in the point \(\frac12(a+b+c+d)\)

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2012 Paper 1 Q6
D: 1516.0 B: 1484.0
trigonometry angle of elevation cotangent trigonometric identity inequality geometric proof flagpole circle geometry product-to-sum formulas

A thin circular path with diameter \(AB\) is laid on horizontal ground. A vertical flagpole is erected with its base at a point \(D\) on the diameter \(AB\). The angles of elevation of the top of the flagpole from \(A\) and \(B\) are \(\alpha\) and \(\beta\) respectively (both are acute). The point \(C\) lies on the circular path with \(DC\) perpendicular to \(AB\) and the angle of elevation of the top of the flagpole from \(C\) is \(\phi\). Show that \(\cot\alpha\cot \beta = \cot^2\phi\). Show that, for any \(p\) and \(q\), \[ \cos p \cos q \sin^2\tfrac12(p+q) - \sin p\sin q \cos^2 \tfrac12 (p+q) = \tfrac12 \cos(p+q) -\tfrac12 \cos(p+q)\cos(p-q) .\] Deduce that, if \(p\) and \(q\) are positive and \( p+q \le \tfrac12 \pi\), then \[ \cot p\cot q\, \ge \cot^2 \tfrac12(p+q) \, \] and hence show that \(\phi \le \tfrac12(\alpha+\beta)\) when \( \alpha +\beta \le \tfrac12 \pi\,\).

Solution:

TikZ diagram
\begin{align*} && \cot \alpha &= \frac{AD}{h} \\ && \cot \beta &= \frac{BD}{h} \\ && \cot \phi &= \frac{DC}h \\ && CD^2 &= AB \cdot BD \tag{intersecting chords} \\ \Rightarrow && \cot^2 \phi &= \cot \alpha \cot \beta \end{align*} \begin{align*} && LHS &= \cos p \cos q \sin^2\tfrac12(p+q) - \sin p\sin q \cos^2 \tfrac12 (p+q) \\ &&&= \cos p \cos q \left ( \frac{1-\cos(p+q)}{2} \right) - \sin p\sin q \left (\frac{1+\cos(p+q)}{2} \right) \\ &&&= \frac12 \left (\cos p \cos q(1-\cos(p+q)) - \sin p\sin q (1+\cos(p+q)) \right) \\ &&&= \frac12 \left ((\cos p \cos q- \sin p\sin q) - (\cos p \cos q+ \sin p\sin q)\cos(p+q) \right) \\ &&&= \frac12 \left (\cos(p+q) - \cos (p-q)\cos(p+q) \right) \\ &&&= RHS \end{align*} Therefore \begin{align*} \cot p \cot q -\cot^2 \tfrac12 (p+q) &= \frac{\tfrac12 \cos(p+q) -\tfrac12 \cos(p+q)\cos(p-q)}{\sin p \sin q \sin^2 \tfrac12(p+q)} \\ &=\frac{\cos(p+q)(1-\cos(p-q))}{\sin p \sin q \sin^2 \tfrac12(p+q)} \end{align*} Since \(p,q\) are acute, the denominator is positive. Since \(p+q \leq \frac{\pi}{2}\), we have \(\cos(p+q) \geq 0\). Also \((1-\cos(p-q)) \geq 0\). Thus, the expression is \(\geq 0\). So we must have \begin{align*} && \cot^2 \phi &= \cot \alpha \cot \beta \\ &&&\geq \cot^2 \tfrac12(\alpha+\beta) \end{align*} Since \(\cot\) is decreasing on \((0, \tfrac12 \pi)\) we can deduce \(\phi \leq \tfrac12 (\alpha+\beta)\)

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2011 Paper 1 Q3
D: 1500.0 B: 1500.0
trigonometry trig identity product-to-sum differentiation cotangent cosecant triple angle formula proof

Prove the identity \[ 4\sin\theta \sin(\tfrac13\pi-\theta) \sin (\tfrac13\pi+\theta)= \sin 3\theta\, . \tag{\(*\)}\]

  1. By differentiating \((*)\), or otherwise, show that \[ \cot \tfrac19\pi - \cot \tfrac29\pi + \cot \tfrac49\pi = \sqrt3\,. \]
  2. By setting \(\theta = \frac16\pi-\phi\) in \((*)\), or otherwise, obtain a similar identity for \(\cos3\theta\) and deduce that \[ \cot \theta \cot (\tfrac13\pi-\theta) \cot (\tfrac13\pi+\theta) =\cot3\theta\,. \] Show that \[ \cosec \tfrac19\pi -\cosec \tfrac59\pi +\cosec \tfrac79\pi = 2\sqrt3\,. \]

Solution: \begin{align*} && LHS &= 4\sin\theta \sin(\tfrac13\pi-\theta) \sin (\tfrac13\pi+\theta) \\ &&&= 4 \sin \theta \left (\tfrac{\sqrt{3}}{2}\cos \theta - \tfrac12 \sin \theta \right)\left (\tfrac{\sqrt{3}}{2}\cos \theta + \tfrac12 \sin \theta \right) \\ &&&= 4 \sin \theta \left (\tfrac{3}{4}\cos^2 \theta - \tfrac14 \sin^2 \theta \right) \\ &&&= 3\sin \theta - 4\sin^3 \theta \\ &&&= \cos 3 \theta = RHS \end{align*}

  1. \(\,\) \begin{align*} && 3 \cos 3 \theta &= \sin 3 \theta \left (\cot \theta - \cot (\tfrac13\pi - \theta) + \cot (\tfrac13\pi + \theta) \right) \\ \Rightarrow && 3 \cot 3\theta &= \cot \theta - \cot (\tfrac13\pi - \theta) + \cot (\tfrac13\pi + \theta) \\ \theta = \tfrac{\pi}{9}: && 3\cot \frac{\pi}{3} &= \cot \tfrac{\pi}{9} - \cot \tfrac{2}{9}\pi + \cot \tfrac49 \pi \\ \Rightarrow && \sqrt{3} &= \cot \tfrac{\pi}{9} - \cot \tfrac{2}{9}\pi + \cot \tfrac49 \pi \end{align*}
  2. \(\,\) \begin{align*} \theta = \tfrac16\pi - \phi && \sin(\tfrac12\pi - 3\phi) &= 4\sin(\tfrac16\pi - \phi)\sin(\phi+\tfrac16\pi)\sin(\tfrac12\pi - \phi) \\ \Rightarrow && \cos 3\phi &= 4\cos(\phi - \tfrac13\pi)\cos(\tfrac13\pi - \phi)\cos\phi \\ \Rightarrow && \cot 3\theta &= \cot \theta\cot(\phi - \tfrac13\pi)\cot(\tfrac13\pi - \phi) \tag{dividing by (\(*\))} \\ \\ \frac{\d}{\d \theta}:&& -\csc^2 3\phi &= \cot3\phi \left (-\csc^2 \phi\tan \phi+\csc^2 (\tfrac13\pi - \phi) \tan (\tfrac13\pi - \phi) -\csc^2(\phi - \tfrac13\pi)\tan (\phi - \tfrac13\pi) \right) \\ \Rightarrow && \csc^23\phi\tan3\phi & = 2( \csc2\phi- \csc(\tfrac{2}{3}\pi - 2\phi)+\csc(\phi - \tfrac23\pi)) \\ \phi = \frac{1}{18}\pi: && 4\sqrt{3} &= 2(\csc \tfrac{1}{9}\pi - \csc \tfrac59\pi + \csc \tfrac79 \pi) \\ \end{align*} and the result follows.

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2005 Paper 1 Q7
D: 1500.0 B: 1516.0
telescoping product product notation trigonometry simplification cotangent partial fractions sine rule compound angle formula

The notation \(\displaystyle \prod^n_{r=1} \f (r)\) denotes the product $\f (1) \times \f (2) \times \f(3) \times \cdots \times \f(n)$. %For example, \(\displaystyle \prod_{r=1}^4 r = 24\). %Simplify \(\displaystyle \prod^n_{r=1} \frac{\g (r) }{ \g (r-1) }\). %You may assume that \(\g (r) \neq 0\) for any integer \(0 \le r \le n \). Simplify the following products as far as possible:

  1. \(\displaystyle \prod^n_{r=1} \l \frac{r+ 1 }{ r } \r\,\);
  2. \(\displaystyle \prod^n_{r=2} \l \frac{r^2 -1}{r^2 } \r\,\);
  3. $\displaystyle \prod^n_{r=1} \l {\cos \frac{2\pi }{ n} + \sin \frac{2\pi}{ n} \cot \frac{\l 2r-1 \r \pi }{ n} }\r\,$, where \(n\) is even.

Solution:

  1. \(\,\) \begin{align*} \prod^n_{r=1} \left ( \frac{r+ 1 }{ r } \right) &= \frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdots \frac{n-1}{n-2} \cdot \frac{n}{n-1} \cdot \frac{n+1}{n} \\ &= \frac{n+1}{1} = n+1 \end{align*}
  2. \(\,\) \begin{align*} \prod^n_{r=2} \left ( \frac{r^2 -1}{r^2 } \right) &= \prod^n_{r=2} \left ( \frac{(r -1)(r+1)}{r^2 } \right) \\ &= \left ( \frac{1}{2} \cdot \frac{3}{2} \right) \cdot \left ( \frac{2}{3} \cdot \frac{4}{3} \right) \cdots \left ( \frac{r-1}{r} \cdot \frac{r+1}{r}\right) \cdots \frac{n-1}{n} \cdot \frac{n+1}{n} \\ &= \frac{1}{n} \cdot \frac{n+1}{2} \\ &= \frac{n+1}{2n} \end{align*}
  3. When \(n\) is odd, the product is undefined, since we have a \(\cot \pi\) lurking in there. \begin{align*} \prod^n_{r=1} \left ( {\cos \frac{2\pi }{ n} + \sin \frac{2\pi}{ n} \cot \frac{ (2r-1 ) \pi }{ n} } \right) &= \prod^n_{r=1} \left ( {\cos \frac{2\pi }{ n} + \sin \frac{2\pi}{ n} \frac{\cos \frac{ (2r-1 ) \pi }{ n}}{\sin\frac{ (2r-1 ) \pi }{ n}} } \right) \\ &= \prod^n_{r=1} \frac{1}{\sin\frac{ (2r-1 ) \pi }{ n}} \left ( {\cos \frac{2\pi }{ n} \sin\frac{ (2r-1 ) \pi }{ n} + \sin \frac{2\pi}{ n} \cos \frac{ (2r-1 ) \pi }{ n} } \right) \\ &= \prod^n_{r=1} \frac{1}{\sin\frac{ (2r-1 ) \pi }{ n}} \sin \left ( \frac{2\pi}{n} + \frac{(2r-1)\pi}{n} \right) \\ &= \prod^n_{r=1} \frac{1}{\sin\frac{ (2r-1 ) \pi }{ n}} \sin \left ( \frac{(2r+1)\pi}{n} \right) \\ &= \frac{\sin \frac{3\pi}{n}}{\sin \frac{\pi}{n}} \cdot \frac{\sin \frac{5\pi}{n}}{\sin \frac{3\pi}{n}} \cdots \frac{\sin \frac{(2n+1)\pi}{n}}{\sin \frac{(2n-1)\pi}{n}} \\ &= \frac{\sin \frac{(2n+1)\pi}{n}}{\sin \frac{\pi}{n}} \\ &= 1 \end{align*}

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1989 Paper 2 Q2
D: 1600.0 B: 1543.0
Taylor series trigonometric power series coefficients cotangent cosecant tangent recurrence relation identity manipulation

Let \begin{alignat*}{2} \tan x & =\ \ \, \quad{\displaystyle \sum_{n=0}^{\infty}a_{n}x^{n}} & & \text{ for small }x,\\ x\cot x & =1+\sum_{n=1}^{\infty}b_{n}x^{n}\quad & & \text{ for small }x\text{ and not zero}. \end{alignat*} Using the relation \[ \cot x-\tan x=2\cot2x,\tag{*} \] or otherwise, prove that \(a_{n-1}=(1-2^{n})b_{n}\), for \(n\geqslant1\). Let \[ x\mathrm{cosec}x=1+{\displaystyle \sum_{n=1}^{\infty}c_{n}x^{n}\quad\text{ for small }x\neq0. \qquad \qquad \, } \] Using a relation similar to \((*)\) involving \(2\mathrm{cosec}2x\), or otherwise, prove that \[ c_{n}=\frac{2^{n-1}-1}{2^{n}-1}\frac{1}{2^{n-1}}a_{n-1}\qquad(n\geqslant1). \]

Solution: \begin{align*} && \cot x - \tan x &= 2 \cot 2x \\ \Rightarrow && x\cot x - x\tan x &= 2x\cot 2x \\ \Rightarrow && 1 + \sum_{n=1}^{\infty} b_n x^n - \sum_{n=0}^{\infty}a_n x^{n+1} &= 1 + \sum_{n=1}^{\infty} b_n (2x)^n \\ \Rightarrow && \sum_{n=1}^{\infty}(1-2^n)b_nx^n &= \sum_{n=1}^{\infty} a_{n-1}x^n \\ \Rightarrow && a_{n-1} &= (1-2^n)b_n \quad \text{if }n \geq 1 \end{align*} \begin{align*} \cot x + \tan x &= 2 \cosec 2x \end{align*} So \begin{align*} && \cot x + \tan x &= 2 \cosec 2x \\ \Rightarrow && x \cot x + x\tan x &= 2x \cosec 2x \\ \Rightarrow && 1 + \sum_{n=1}^{\infty} b_n x^n + \sum_{n=0}^{\infty} a_n x^{n+1} &= 1+\sum_{n=1}^\infty c_n (2x)^n \\ \Rightarrow && \sum_{n=1}^{\infty} \frac{1}{1-2^n}a_{n-1} +\sum_{n=1}^{\infty}a_{n-1}x^n &= \sum_{n=1}^{\infty} 2^nc_n x^n \\ \Rightarrow && c_n &= \frac{1}{2^n} \left ( 1 + \frac{1}{1-2^n} \right)a_{n-1} \\ &&&= \frac1{2^n} \frac{2^n-2}{2^n-1} a_{n-1}\\ &&&= \frac1{2^{n-1}}\frac{2^{n-1}-1}{2^n-1} a_{n-1} \end{align*}

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