OCR C3 — Question 9 13 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAddition & Double Angle Formulae
TypeDerive triple angle then solve equation
DifficultyChallenging +1.2 Part (i) is a standard trigonometric identity derivation using compound angle formulas (routine for C3). Part (ii) requires recognizing the triple angle formula from (i) and finding maximum value, which is moderately challenging. Part (iii) involves product-to-sum formulas and solving a more complex trigonometric equation with restricted domain, requiring several non-trivial steps. Overall, this is harder than average C3 questions due to the multi-step reasoning and connections between parts, but still within standard A-level techniques.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals

  1. By first writing \(\sin 3\theta\) as \(\sin(2\theta + \theta)\), show that $$\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta.$$ [4]
  2. Determine the greatest possible value of $$9 \sin(\frac{10}{3}\alpha) - 12 \sin^3(\frac{10}{3}\alpha),$$ and find the smallest positive value of \(\alpha\) (in degrees) for which that greatest value occurs. [3]
  3. Solve, for \(0° < \beta < 90°\), the equation \(3 \sin 6\beta \cos 2\beta = 4\). [6]

\begin{enumerate}[label=(\roman*)]
\item By first writing $\sin 3\theta$ as $\sin(2\theta + \theta)$, show that
$$\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta.$$ [4]

\item Determine the greatest possible value of
$$9 \sin(\frac{10}{3}\alpha) - 12 \sin^3(\frac{10}{3}\alpha),$$
and find the smallest positive value of $\alpha$ (in degrees) for which that greatest value occurs. [3]

\item Solve, for $0° < \beta < 90°$, the equation $3 \sin 6\beta \cos 2\beta = 4$. [6]
\end{enumerate}

\hfill \mbox{\textit{OCR C3  Q9 [13]}}