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1998 Paper 1 Q6
D: 1500.0 B: 1500.0
trigonometry recurrence relation half-angle formula product of cosines sequences proof by induction cos x arithmetic-geometric mean

Let \(a_{1}=\cos x\) with \(0 < x < \pi/2\) and let \(b_{1}=1\). Given that \begin{eqnarray*} a_{n+1}&=&{\textstyle \frac{1}{2}}(a_{n}+b_{n}),\\[2mm] b_{n+1}&=&(a_{n+1}b_{n})^{1/2}, \end{eqnarray*} find \(a_{2}\) and \(b_{2}\) and show that \[a_{3}=\cos\frac{x}{2}\cos^{2}\frac{x}{4} \ \quad\mbox{and}\quad \ b_{3}=\cos\frac{x}{2}\cos\frac{x}{4}.\] Guess general expressions for \(a_{n}\) and \(b_{n}\) (for \(n\ge2\)) as products of cosines and verify that they satisfy the given equations.

Solution: \begin{array}{c|c|c} n & a_n & b_n \\ \hline 1 & \cos x & 1 \\ \hline 2 & \frac12(1 + \cos x) & \sqrt{a_2} \\ &=\frac12(1+2\cos^2 \frac{x}{2}-1)& \sqrt{a_2} \\ &= \cos^2 \frac{x}{2} & \cos \frac{x}{2} \\ \hline 3 & \frac12(\cos^2 \frac{x}{2}+\cos \frac{x}{2}) & \sqrt{a_3\cos \frac{x}{2}} \\ &= \cos \frac{x}{2} \cdot \frac12 (\cos \frac{x}{2}+1) & \sqrt{a_3\cos \frac{x}{2}} \\ &= \cos \frac{x}{2} \cos^2 \frac{x}{4} & \cos \frac{x}{2} \cos \frac{x}{4} \end{array} Claim: \(\displaystyle a_n = \cos \frac{x}{2^{n-1}}\prod_{k=1}^{n-1} \cos \frac{x}{2^k}\), \(\displaystyle b_n = \prod_{k=1}^{n-1} \cos \frac{x}{2^k}\) Claim: \(a_{n+1} = \frac12(a_n + b_n)\) Proof: \begin{align*} && \frac12(a_n + b_n) &= \frac12 \left ( \cos \frac{x}{2^{n-1}}\prod_{k=1}^{n-1} \cos \frac{x}{2^k} + \prod_{k=1}^{n-1} \cos \frac{x}{2^k} \right) \\ &&&= \prod_{k=1}^{n-1} \cos \frac{x}{2^k} \frac12\left (\cos \frac{x}{2^{n-1}} + 1 \right) \\ &&&= \left ( \prod_{k=1}^{n-1} \cos \frac{x}{2^k} \right) \cos^{2} \frac{x}{2^n} \\ &&&= \cos \frac{x}{2^n} \prod_{k=1}^{n} \cos \frac{x}{2^k} \\ &= a_{n+1} \end{align*} Claim: \(b_{n+1} = \sqrt{a_{n+1}b_n}\) Proof: \begin{align*} && \sqrt{a_{n+1}b_n} &= \sqrt{ \cos \frac{x}{2^n} \prod_{k=1}^{n} \cos \frac{x}{2^k} \cdot \prod_{k=1}^{n-1} \cos \frac{x}{2^k} }\\ &&&= \prod_{k=1}^{n-1} \cos \frac{x}{2^k} \sqrt{\cos ^2\frac{x}{2^{n}}} \\ &&&= \prod_{k=1}^{n} \cos \frac{x}{2^k} \\ &&&= b_{n+1} \end{align*}

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1994 Paper 1 Q4
D: 1484.0 B: 1628.6
integration trigonometric identity completing the square inverse tangent definite integral t-substitution half-angle formula

Show that

  1. \(\dfrac{1-\cos\alpha}{\sin\alpha}=\tan\frac{1}{2}\alpha,\)
  2. if \(\left|k\right|<1\) then \({\displaystyle \int\frac{\mathrm{d}x}{1-2kx+x^{2}}=\frac{1}{\sqrt{1-k^{2}}}\tan^{-1}\left(\frac{x-k}{\sqrt{1-k^{2}}}\right)+C,}\) where \(C\) is a constant of integration.
Hence, or otherwise, show that if \(0<\alpha<\pi\) then \[ \int_{0}^{1}\frac{\sin\alpha}{1-2x\cos\alpha+x^{2}}\,\mathrm{d}x=\frac{\pi-\alpha}{2}. \]

Solution:

  1. \begin{align*} \frac{1-\cos \alpha}{\sin \alpha} &= \frac{1-(1-2\sin^2 \frac{\alpha}{2})}{2 \sin \frac \alpha2 \cos \frac\alpha2} \\ &= \frac{2 \sin^2 \frac \alpha2}{2 \sin \frac \alpha2 \cos \frac\alpha2} \\ &= \frac{\sin \frac \alpha2}{ \cos \frac\alpha2} \\ &= \tan \tfrac{\alpha}{2} \end{align*}
  2. \begin{align*} \int\frac{\mathrm{d}x}{1-2kx+x^{2}} &= \int \frac{\d x}{(x-k)^2+1-k^2} \\ &= \frac{1}{1-k^2}\int \frac{\d x}{\left (\frac{x-k}{\sqrt{1-k^2}} \right)^2+1} \\ &= \frac{1}{\sqrt{1-k^2}} \tan^{-1} \left (\frac{x-k}{\sqrt{1-k^2}} \right)+C \end{align*}
\begin{align*} \int_{0}^{1}\frac{\sin\alpha}{1-2x\cos\alpha+x^{2}}\,\mathrm{d}x &= \sin \alpha \left [\frac{1}{\sqrt{1-\cos ^2\alpha}} \tan^{-1} \left ( \frac{x - \cos \alpha}{\sqrt{1-\cos^2\alpha}} \right) \right]_0^1 \\ &= \tan^{-1} \left ( \frac{1 - \cos \alpha}{\sin \alpha} \right) -\tan^{-1} \left ( \frac{- \cos \alpha}{\sin \alpha} \right) \\ &= \tan^{-1} \tan \tfrac{\alpha}{2} + \tan^{-1} \cot \alpha \\ &= \frac{\alpha}{2} + \frac{\pi}{2} - \alpha \\ &= \frac{\pi-\alpha}{2} \end{align*}

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