Challenging +1.2 This is a standard Further Maths FP2 question requiring de Moivre's theorem application and algebraic manipulation to derive a multiple angle formula, followed by solving a trigonometric equation. While it involves multiple steps and some algebraic dexterity, it follows a well-established template that FP2 students practice extensively. The proof in part (b) is straightforward once the identity is established, requiring factorization and basic trigonometric reasoning. This is moderately above average difficulty due to the Further Maths content and multi-step nature, but remains a textbook-style exercise rather than requiring novel insight.
6. (a) Use de M oivre's Theorem to show that
$$\sin 5 \theta = 16 \sin ^ { 5 } \theta - 20 \sin ^ { 3 } \theta + 5 \sin \theta .$$
(b) Hence or otherwise, prove that the only real solutions of the equation
$$\sin 5 \theta = 5 \sin \theta ,$$
are given by \(\theta = n \tau\), where \(n\) is an integer.
6. (a) Use de M oivre's Theorem to show that
$$\sin 5 \theta = 16 \sin ^ { 5 } \theta - 20 \sin ^ { 3 } \theta + 5 \sin \theta .$$
(b) Hence or otherwise, prove that the only real solutions of the equation
$$\sin 5 \theta = 5 \sin \theta ,$$
are given by $\theta = n \tau$, where $n$ is an integer.\\
\hfill \mbox{\textit{Edexcel FP2 Q6 [9]}}