OCR FP3 2010 January — Question 7 13 marks

Exam BoardOCR
ModuleFP3 (Further Pure Mathematics 3)
Year2010
SessionJanuary
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeExpress roots in trigonometric form
DifficultyChallenging +1.3 This is a structured FP3 question on de Moivre's theorem with clear scaffolding. Part (i) is routine, part (ii) requires standard expansion of (cos θ + i sin θ)^6 and algebraic manipulation, and part (iii) uses roots of the derived polynomial—a classic technique. While requiring multiple steps and FP3 content (inherently harder), the question guides students through each stage with no novel insight needed, making it moderately above average difficulty.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals4.02q De Moivre's theorem: multiple angle formulae

  1. Solve the equation \(\cos 6\theta = 0\), for \(0 < \theta < \pi\). [3]
  2. By using de Moivre's theorem, show that $$\cos 6\theta \equiv (2\cos^2\theta - 1)(16\cos^4\theta - 16\cos^2\theta + 1).$$ [5]
  3. Hence find the exact value of $$\cos\left(\frac{1}{12}\pi\right)\cos\left(\frac{5}{12}\pi\right)\cos\left(\frac{7}{12}\pi\right)\cos\left(\frac{11}{12}\pi\right),$$ justifying your answer. [5]

\begin{enumerate}[label=(\roman*)]
\item Solve the equation $\cos 6\theta = 0$, for $0 < \theta < \pi$. [3]
\item By using de Moivre's theorem, show that
$$\cos 6\theta \equiv (2\cos^2\theta - 1)(16\cos^4\theta - 16\cos^2\theta + 1).$$ [5]
\item Hence find the exact value of
$$\cos\left(\frac{1}{12}\pi\right)\cos\left(\frac{5}{12}\pi\right)\cos\left(\frac{7}{12}\pi\right)\cos\left(\frac{11}{12}\pi\right),$$
justifying your answer. [5]
\end{enumerate}

\hfill \mbox{\textit{OCR FP3 2010 Q7 [13]}}