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2003 Paper 1 Q3
D: 1500.0 B: 1484.0
trigonometry double angle formula trigonometric equations half-angle general solution factorisation

  1. Show that \( 2\sin(\frac12\theta)=\sin \theta\) if and only if \(\sin(\frac12\theta)=0\,\).
  2. Solve the equation \(2\tan (\frac12\theta) = \tan\theta\,\).
  3. Show that \(2\cos(\frac12\theta)=\cos \theta\) if and only if \(\theta=(4n+2)\pi\pm 2\phi\) where \(\phi\) is defined by \(\cos \phi=\frac12(\sqrt 3-1)\;\), \(0\le \phi\le \frac{1}{2}\pi\), and \(n\) is any integer.

Solution:

  1. \(\,\) \begin{align*} && 2 \sin (\tfrac12 \theta) &= \sin \theta \\ \Leftrightarrow && 2 \sin (\tfrac12 \theta) &= 2\sin (\tfrac12 \theta) \cos (\tfrac12 \theta) \\ \Leftrightarrow && 0 &= 2\sin(\tfrac12\theta)(1-\cos(\tfrac12 \theta)) \\ \Leftrightarrow && 0 = \sin(\tfrac12 \theta) &\text{ or } 1 = \cos(\tfrac12 \theta) \\ \Leftrightarrow && 0 &= \sin(\tfrac12 \theta) \end{align*}
  2. Let \(= \tan(\tfrac12 \theta)\), then \begin{align*} && 2t &= \frac{2t}{1-t^2} \\ \Leftrightarrow && 0 &= \frac{2t(1-(1-t^2)}{1-t^2} \\ &&&= \frac{2t^3}{1-t^2} \\ \Leftrightarrow && t&= 0 \\ \Leftrightarrow && \frac12\theta &= n \pi \\ \Leftrightarrow && \theta &= 2n\pi \end{align*}
  3. Let \(c = \cos(\tfrac12 \theta)\), then \begin{align*} && 2c &= 2c^2 - 1 \\ && 0 &= 2c^2-2c-1 \\ \Leftrightarrow && c &= \frac{2 \pm \sqrt{4+8}}{4} \\ &&&= \frac{1 \pm \sqrt{3}}{2} \\ \Leftrightarrow && c &= \frac{1 - \sqrt{3}}{2} \\ \Leftrightarrow && \frac12 \theta &= \pm \cos^{-1} \frac{1 - \sqrt{3}}{2} + 2n \pi \\ &&&= \mp (\phi+\pi) + 2n \pi \\ \Leftrightarrow && \theta &= (4n+2)\pi \pm 2\phi \end{align*}

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