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2017 Paper 3 Q8
D: 1700.0 B: 1500.0
sequences summation by parts Abel summation trigonometric series telescoping product-to-sum identities cosecant

Prove that, for any numbers \(a_1, a_2, \ldots\,,\) and \(b_1, b_2, \ldots\,,\) and for \(n\ge1\), \[ \sum_{m=1}^n a_m(b_{m+1} -b_m) = a_{n+1}b_{n+1} -a_1b_1 -\sum_{m=1}^n b_{m+1}(a_{m+1} -a_m) \,. \]

  1. By setting \(b_m = \sin mx\), show that \[ \sum_{m=1}^n \cos (m+\tfrac12)x = \tfrac12 \big(\sin (n+1)x - \sin x \big) \cosec \tfrac12 x \,. \] Note: $\sin A - \sin B = \displaystyle 2 \cos \big( \tfrac{{\displaystyle A+B\vphantom{_1}}} {\displaystyle 2\vphantom{^1}} \big)\, \sin\big( \tfrac{{\displaystyle A-B\vphantom{_1}}}{\displaystyle 2\vphantom{^1}} \big)\, $.
  2. Show that \[ \sum_{m=1}^n m\sin mx = \big (p \sin(n+1)x +q \sin nx\big) \cosec^2 \tfrac12 x \,, \] where \(p\) and \(q\) are to be determined in terms of \(n\). Note: \(2\sin A \sin B = \cos (A-B) - \cos (A+B)\,\); Note: \(2\cos A \sin B = \sin (A+B) - \sin (A-B)\,\).

Solution: \begin{align*} \sum_{m=1}^n a_m(b_{m+1} -b_m) +\sum_{m=1}^n b_{m+1}(a_{m+1} -a_m) &= \sum_{m=1}^n \left (a_{m+1}b_{m+1}-a_mb_m \right) \\ &= a_{n+1}b_{n+1} - a_1b_1 \end{align*} And the result follows.

  1. Let \(b_m = \sin m x \), \(a_m = \cosec \frac{x}{2}\), so \begin{align*} && \sum_{m=1}^n \cosec \frac{x}{2} \left (\sin (m+1)x - \sin mx \right) &= \sum_{m=1}^n \cosec \frac{x}{2} 2 \cos \left ( \frac{2m+1}{2}x \right) \sin \left ( \frac{(m+1)-m}{2}x \right) \\ &&&=2 \sum_{m=1}^n\cos \left ( (m + \tfrac12)x \right)\\ \\ \Rightarrow && \sum_{m=1}^n\cos \left ( (m + \tfrac12)x \right) &= \tfrac12 \cosec \tfrac{x}{2}\left ( \sin(n+1)x - \sin x \right) \end{align*}
  2. \(\,\) \begin{align*} && b_{m+1}-b_m &= \sin m x \sin \tfrac12 x \\ &&&= \frac12 \left ( \cos (m-\tfrac12)x - \cos (m+\tfrac12)x \right)\\ \Rightarrow && b_m &= -\tfrac12 \cos (m - \tfrac12)x\\ && a_m &= m \\ \Rightarrow && \sum_{m=1}^n m \sin m x \sin \tfrac12 x &= (n+1) b_{n+1} - 1 \cdot b_1 - \sum_{m=1}^n b_{m+1} \cdot 1 \\ &&&= -(n+1) \tfrac12\cos(n+1-\tfrac12)x+\tfrac12\cos(\tfrac12x) + \tfrac12\sum_{m=1}^n \cos(m+\tfrac12)x \\ &&&= -(n+1) \tfrac12\cos(n+1-\tfrac12)x+\tfrac12\cos(\tfrac12x) + \tfrac14 \cosec \tfrac{x}{2}\left ( \sin(n+1)x - \sin x \right) \\ &&&= -(n+1) \tfrac12\cos(n+1-\tfrac12)x+ \tfrac14 \cosec \tfrac{x}{2}\sin(n+1)x \\ &&&= \tfrac12\cosec\tfrac{x}2 \left (\tfrac12 \sin (n+1)x-(n+1)\cos(n+\tfrac12)x\sin\tfrac12x \right) \\ &&&= \tfrac12\cosec\tfrac{x}2 \left (\tfrac12 \sin (n+1)x-(n+1)\tfrac12 \left ( \sin (n+1)x - \sin nx \right) \right) \\ &&&= \tfrac14 \cosec \tfrac{x}{2} \left ( -n \sin (n+1)x +(n+1) \sin n x \right) \end{align*} Therefore \(p = -\frac{n}4, q = \frac{n+1}{4}\)
Notice the connection here to integration by parts.

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

  1. Deduce that, when \(n\) is large, \[ \sum_{k=0}^n \sin \left(\frac{k\pi}{n}\right) \approx \frac{2n}\pi\;. \]
  2. 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*}

  1. 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*}
  2. \(\,\) \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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1990 Paper 3 Q4
D: 1700.0 B: 1516.0
complex numbers geometric series binomial theorem trigonometric series De Moivre summation real and imaginary parts

Given that \(\sin\beta\neq0,\) sum the series \[ \cos\alpha+\cos(\alpha+2\beta)+\cdots+\cos(\alpha+2r\beta)+\cdots+\cos(\alpha+2n\beta) \] and \[ \cos\alpha+\binom{n}{1}\cos(\alpha+2\beta)+\cdots+\binom{n}{r}\cos(\alpha+2r\beta)+\cdots+\cos(\alpha+2n\beta). \] Given that \(\sin\theta\neq0,\) prove that \[ 1+\cos\theta\sec\theta+\cos2\theta\sec^{2}\theta+\cdots+\cos r\theta\sec^{r}\theta+\cdots+\cos n\theta\sec^{n}\theta=\frac{\sin(n+1)\theta\sec^{n}\theta}{\sin\theta}. \]

Solution: \begin{align*} \sum_{r = 0}^n \cos (\alpha + 2r \beta) &= \sum_{r = 0}^n \textrm{Re} \left ( \exp(i(\alpha + 2r \beta)) \right) \\ &= \textrm{Re} \left (\sum_{r = 0}^n \exp(i(\alpha + 2r \beta)) \right) \\ &= \textrm{Re} \left (e^{i \alpha} \sum_{r = 0}^n \ (e^{i 2 \beta})^r\right) \\ &= \textrm{Re} \left (e^{i \alpha} \frac{e^{2(n+1)\beta i}-1}{e^{2\beta i}-1} \right) \\ &= \textrm{Re} \left (e^{i \alpha} \frac{e^{(n+1)\beta i} (e^{(n+1)\beta i}-e^{-(n+1)\beta i})}{e^{\beta i}(e^{\beta i}-e^{-\beta i})} \right) \\ &= \textrm{Re} \left (\frac{e^{i \alpha} e^{(n+1)\beta i}}{e^{\beta i}} \frac{\sin (n+1) \beta}{\sin \beta} \right) \\ &= \textrm{Re} \left ( e^{i(\alpha + n \beta)}\frac{\sin (n+1) \beta}{\sin \beta} \right) \\ &= \frac{\cos (\alpha + n \beta)\sin (n+1) \beta}{\sin \beta} \end{align*} \begin{align*} \sum_{r = 0}^n \binom{n}{r} \cos (\alpha + 2r \beta) &= \sum_{r = 0}^n \textrm{Re} \left ( \binom{n}{r}\exp(i(\alpha + 2r \beta)) \right) \\ &= \textrm{Re} \left (\sum_{r = 0}^n \binom{n}{r} \exp(i(\alpha + 2r \beta)) \right) \\ &= \textrm{Re} \left (e^{i \alpha}(e^{2\beta i}+1)^n \right) \\ &= \textrm{Re} \left (e^{i \alpha}e^{n\beta i}(e^{\beta i}+e^{-\beta i})^n \right) \\ &= \textrm{Re} \left (e^{i \alpha}e^{n\beta i}2^n \cos^n \beta \right) \\ &= 2^n \cos(\alpha + n \beta) \cos ^n \beta \end{align*} \begin{align*} \sum_{r = 0}^n \cos r \theta \sec^r \theta &= \sum_{r = 0}^n \textrm{Re} ( e^{i r \theta})\sec^r \theta \\ &= \textrm{Re} \left ( \sum_{r=0}^n e^{i r \theta} \sec^r \theta\right) \\ &= \textrm{Re} \left ( \frac{e^{i (n+1) \theta}\sec^{n+1} \theta -1}{e^{i \theta}\sec \theta -1} \right) \\ &= \textrm{Re} \left ( \frac{e^{i (n+1) \theta}\sec^{n} \theta -\cos \theta}{e^{i \theta} -\cos \theta} \right) \\ &= \textrm{Re} \left ( \frac{e^{i (n+1) \theta}\sec^{n} \theta -\cos \theta}{i \sin \theta} \right) \\ &= \frac{1}{\sin \theta} \textrm{Im} \left ( e^{i (n+1) \theta}\sec^{n} \theta -\cos \theta \right) \\ &= \frac{\sin(n+1) \theta \sec^{n} \theta}{\sin \theta} \end{align*}

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1987 Paper 1 Q7
D: 1500.0 B: 1500.0
complex numbers geometric series trigonometric series summation De Moivre imaginary part tangent

Sum each of the series \[ \sin\left(\frac{2\pi}{23}\right)+\sin\left(\frac{6\pi}{23}\right)+\sin\left(\frac{10\pi}{23}\right)+\cdots+\sin\left(\frac{38\pi}{23}\right)+\sin\left(\frac{42\pi}{23}\right) \] and \[ \sin\left(\frac{2\pi}{23}\right)-\sin\left(\frac{6\pi}{23}\right)+\sin\left(\frac{10\pi}{23}\right)-\cdots-\sin\left(\frac{38\pi}{23}\right)+\sin\left(\frac{42\pi}{23}\right), \] giving each answer in terms of the tangent of a single angle. {[}No credit will be given for a numerical answer obtained purely by use of a calculator.{]}

Solution: \(\sin x = \frac{e^{ix} - e^{-ix}}{2i}\). Also let \(z = e^{ \frac{2\pi i}{23}}\) \begin{align*} \sum_{k=0}^{10} \sin \l \frac{(4k +2)\pi}{23} \r &= \sum_{k=0}^{10} \textrm{Im} \l \exp\l \frac{(4k +2)\pi i}{23} \r \r \\ &= \textrm{Im} \l \sum_{k=0}^{10} \exp\l \frac{(4k +2)\pi i}{23} \r \r \\ &= \textrm{Im} \l e^{ \frac{2\pi i}{23}} \sum_{k=0}^{10} z^{2k} \r \\ &= \textrm{Im} \l z \l \frac{z^{22}-1}{z^2-1} \r \r \\ &= \textrm{Im} \l z \l \frac{z^{11}(z^{11}-z^{-11})}{z(z-z^{-1})} \r \r \\ &= \textrm{Im} \l \frac{z^{11}2i \sin \frac{22 \pi}{23} }{2i \sin \frac{2 \pi}{23}} \r \r \\ &= \frac{\sin \frac{22 \pi}{23}}{\sin \frac{2 \pi}{23}} \textrm{Im} ( z^{11}) \\ &= \frac{\sin^2 \frac{22 \pi}{23}}{\sin \frac{2 \pi}{23}} \\ &= \frac{\sin^2 \frac{\pi}{23}}{2\sin \frac{\pi}{23}\cos \frac{\pi}{23}} \\ &= \frac12 \tan \frac{\pi}{23} \end{align*} Similarly, \begin{align*} \sum_{k=0}^{10} (-1)^k\sin \l \frac{(4k +2)\pi}{23} \r &= \sum_{k=0}^{10} \textrm{Im} \l (-1)^k\exp\l \frac{(4k +2)\pi i}{23} \r \r \\ &= \textrm{Im} \l \sum_{k=0}^{10} (-1)^k\exp\l \frac{(4k +2)\pi i}{23} \r \r \\ &= \textrm{Im} \l e^{ \frac{2\pi i}{23}} \sum_{k=0}^{10} (-1)^kz^{2k} \r \\ &= \textrm{Im} \l z \l \frac{z^{22}+1}{z^2+1} \r \r \\ &= \textrm{Im} \l z \l \frac{z^{11}(z^{11}+z^{-11})}{z(z+z^{-1})} \r \r \\ &= \textrm{Im} \l \frac{z^{11}2 \cos \frac{22 \pi}{23} }{2 \cos\frac{2 \pi}{23}} \r \r \\ &= \frac{\cos\frac{22 \pi}{23}}{\cos \frac{2 \pi}{23}} \textrm{Im} ( z^{11}) \\ &= \frac{\cos \frac{22 \pi}{23}\sin \frac{22 \pi}{23}}{\cos\frac{2 \pi}{23}} \\ &= \frac12 \frac{\sin \frac{44 \pi}{23}}{\cos\frac{2 \pi}{23}} \\ &= \frac12 \frac{-\sin \frac{2\pi}{23}}{\cos\frac{2 \pi}{23}} \\ &= -\frac12 \tan \frac{2\pi}{23} \end{align*}

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