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2018 Paper 1 Q6
D: 1500.0 B: 1500.0
trigonometry product-to-sum formula telescoping sum numerical integration midpoint rule trapezium rule area approximation sin x

Use the identity \[ 2 \sin P\,\sin Q = \cos(Q-P)-\cos(Q+P)\, \] to show that \[ 2\sin\theta \,\big (\sin\theta + \sin 3\theta + \cdots + \sin (2n-1)\theta\,\big ) = 1-\cos 2n\theta \,. \]

  1. Let \(A_n\) be the approximation to the area under the curve \(y=\sin x\) from \(x=0\) to \(x=\pi\), using \(n\) rectangular strips each of width \(\frac{{\displaystyle \pi}}{\displaystyle n}\), such that the midpoint of the top of each strip lies on the curve. Show that \[ A_n \sin \left( \frac{\pi}{2n} \right) = \frac \pi n\,. \]
  2. Let \(B_n\) be the approximation to the area under the curve \(y=\sin x\) from \(x=0\) to \(x=\pi\,\), using the trapezium rule with \(n\) strips each of width \(\frac{\displaystyle \pi}{ \displaystyle n}\). Show that \[B_n \sin \left( \frac{\pi}{2n} \right) = \frac{\pi}{n} \cos \left( \frac{\pi}{2n} \right) . \]
  3. Show that \[ \frac{1}{2}(A_n + B_n) = B_{2n}\,, \] and that \[ A_n B_{2n} = A^2_{2n}\, . \]

Solution: \begin{align*} && 2\sin\theta \,\big (\sin\theta + \sin 3\theta + \cdots + \sin (2n-1)\theta\,\big ) &= 2\sin\theta\sin\theta + 2\sin\theta\sin 3\theta + \cdots + 2\sin\theta\sin (2n-1)\theta \\ &&&= \cos((1-1)\theta) - \cos((1+1)\theta)+\cos((3-1)\theta)-\cos((3+1)\theta) + \cdots + \cos (((2n-1)-1)\theta) -\cos(((2n-1)+1)\theta) \\ &&&= \cos 0 - \cos(2n\theta) \\ &&&= 1 - \cos 2n \theta \end{align*}

  1. \(\,\)
    TikZ diagram
    Therefore the area is: \begin{align*} A_n &= \frac{\pi}{n} \sin \left ( \frac{\pi}{2n} \right) + \frac{\pi}{n} \sin \left ( \frac{3\pi}{2n} \right) + \frac{\pi}{n} \sin \left ( \frac{5\pi}{2n} \right) + \cdots \frac{\pi}{n} \sin \left ( \frac{(2n-1)\pi}{2n} \right) \\ &= \frac{\pi}{n} \left( \frac{1-\cos \frac{2n \pi}{2n}}{2\sin \frac{\pi}{2n}} \right) \\ &= \frac{\pi}{n} \frac{1}{\sin \frac{\pi}{2n}} \end{align*} as required
  2. TikZ diagram
    Therefore the area is: \begin{align*} && B_n &= \frac{\pi}{n} \frac{\sin(0)+\sin(\frac{\pi}{n})}{2}+\frac{\pi}{n} \frac{\sin(\frac{\pi}n)+\sin(\frac{2\pi}{n})}{2} + \cdots \frac{\pi}{n} \frac{\sin(\frac{(n-1)\pi}{n})+\sin(\frac{n\pi}{n})}{2} \\ &&&= \frac{\pi}{n} \left ( \sin \frac{\pi}{n} + \sin \frac{2\pi}{n} + \cdots+\sin \frac{(n-1)\pi}{n} \right) \\ \Rightarrow && 2\sin \left ( \frac{\pi}{2n} \right)B_n &= \frac{\pi}{2} \left (2\sin \left ( \frac{\pi}{2n} \right)\sin \frac{\pi}{n} + 2\sin \left ( \frac{\pi}{2n} \right)\sin \frac{2\pi}{n} + \cdots+2\sin \left ( \frac{\pi}{2n} \right)\sin \frac{(n-1)\pi}{n} \right) \\ &&&= \frac{\pi}2 \left ( \cos \frac{\pi}{2n} - \cos \frac{3\pi}{n} + \cos\frac{3\pi}{2n} - \cos \frac{5\pi}{2n} + \cos \frac{(2n-3)\pi}{2n} - \cos \frac{(2n-1)\pi}{2n} \right) \\ &&&= \frac{\pi}{n} \left ( \cos \frac{\pi}{2n} - \cos \left ( \pi - \frac{\pi}{2n} \right) \right) \\ &&&= 2 \frac{\pi}{n} \cos \frac{\pi}{2n} \\ \Rightarrow && \sin \left ( \frac{\pi}{2n} \right)B_n &= \frac{\pi}n \cos \frac{\pi}{2n} \end{align*} as required
  3. \begin{align*} \frac12(A_n+B_n) &= \frac12 \frac{\pi}{n} \frac{1}{\sin \frac{\pi}{2n}} \left ( 1 + \cos \frac{\pi}{2n} \right) \\ &= \frac{\pi}{n}\frac1{2 \sin \frac{\pi}{2n}} \left (2 \cos^2 \frac{\pi}{4n} \right) \\ &=\frac{\pi}{n} \frac{1}{4 \sin \frac{\pi}{4n} \cos \frac{\pi}{4n}} \left (2 \cos^2 \frac{\pi}{4n} \right) \\ &= \frac{\pi}{2n} \frac{\cos \frac{\pi}{4n}}{\sin \frac{\pi}{4n}} \\ &= B_{2n} \\ \\ A_nB_{2n} &= \frac{\pi}{n\sin \frac{\pi}{2n}} \cdot \frac{\pi}{2n} \frac{\cos \frac{\pi}{4n}}{\sin \frac{\pi}{4n}} \\ &= \frac{\pi^2}{(2n)^2} \frac{\cos \frac{\pi}{4n}}{\sin^2 \frac{\pi}{4n} \cos \frac{\pi}{4n}} \\ &= \left ( \frac{\pi}{2n} \frac{1}{\sin \frac{\pi}{4n}}\right)^2 \\ &= A_{2n}^2 \end{align*}

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