| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Solve equations using trigonometric identities |
| Difficulty | Challenging +1.2 This is a Further Maths FP3 question involving de Moivre's theorem and trigonometric identities. Part (i) is routine verification and graphical analysis. Part (ii) requires systematic application of de Moivre's theorem to derive an identity—a standard FP3 technique but requiring careful algebraic manipulation over 5 marks. Part (iii) involves solving a quartic equation that factors nicely and interpreting which solution corresponds to which angle. While this requires multiple techniques and extended working, it follows well-established FP3 patterns without requiring novel insight. The 11-mark total and multi-step nature place it above average difficulty, but it remains a standard Further Maths exercise rather than a problem requiring creative approaches. |
| Spec | 1.05o Trigonometric equations: solve in given intervals4.02q De Moivre's theorem: multiple angle formulae |
\begin{enumerate}[label=(\roman*)]
\item
\begin{enumerate}[label=(\alph*)]
\item Verify, without using a calculator, that $\theta = \frac{1}{8}\pi$ is a solution of the equation $\sin 6\theta = \sin 2\theta$. [1]
\item By sketching the graphs of $y = \sin 6\theta$ and $y = \sin 2\theta$ for $0 \leqslant \theta \leqslant \frac{1}{2}\pi$, or otherwise, find the other solution of the equation $\sin 6\theta = \sin 2\theta$ in the interval $0 < \frac{1}{2}\pi$. [2]
\end{enumerate}
\item Use de Moivre's theorem to prove that
$$\sin 6\theta \equiv \sin 2\theta(16\cos^4 \theta - 16\cos^2 \theta + 3).$$ [5]
\item Hence show that one of the solutions obtained in part (i) satisfies $\cos^2 \theta = \frac{1}{4}(2 - \sqrt{2})$, and justify which solution it is. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 Q7 [11]}}