OCR FP2 2008 June — Question 8

Exam BoardOCR
ModuleFP2 (Further Pure Mathematics 2)
Year2008
SessionJune
TopicPolar coordinates

8 The equation of a curve, in polar coordinates, is $$r = 1 - \sin 2 \theta , \quad \text { for } 0 \leqslant \theta < 2 \pi$$

  1. \includegraphics[max width=\textwidth, alt={}, center]{63a316f6-1c18-4224-930f-0b58112c9f71-3_268_796_1567_717} The diagram shows the part of the curve for which \(0 \leqslant \theta \leqslant \alpha\), where \(\theta = \alpha\) is the equation of the tangent to the curve at \(O\). Find \(\alpha\) in terms of \(\pi\).
  2. (a) If \(\mathrm { f } ( \theta ) = 1 - \sin 2 \theta\), show that \(\mathrm { f } \left( \frac { 1 } { 2 } ( 2 k + 1 ) \pi - \theta \right) = \mathrm { f } ( \theta )\) for all \(\theta\), where \(k\) is an integer.
    (b) Hence state the equations of the lines of symmetry of the curve $$r = 1 - \sin 2 \theta , \quad \text { for } 0 \leqslant \theta < 2 \pi$$
  3. Sketch the curve with equation $$r = 1 - \sin 2 \theta , \quad \text { for } 0 \leqslant \theta < 2 \pi$$ State the maximum value of \(r\) and the corresponding values of \(\theta\).