OCR C3 — Question 5 7 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAddition & Double Angle Formulae
TypeSolve equation with sec/cosec/cot
DifficultyModerate -0.3 Part (i) is pure recall of a standard double angle formula (1 mark). Part (ii) requires finding cos α from sin α using Pythagoras, then substituting into the double angle formula - straightforward but multi-step (3 marks). Part (iii) involves rewriting sec β, using the double angle formula, simplifying to find cos β, then solving - standard technique for C3 but requires careful algebraic manipulation. Overall slightly easier than average due to being mostly procedural application of well-practiced identities.
Spec1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05j 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. Write down the identity expressing \(\sin 2\theta\) in terms of \(\sin \theta\) and \(\cos \theta\). [1]
  2. Given that \(\sin \alpha = \frac{1}{4}\) and \(\alpha\) is acute, show that \(\sin 2\alpha = \frac{1}{8}\sqrt{15}\). [3]
  3. Solve, for \(0° < \beta < 90°\), the equation \(5 \sin 2\beta \sec \beta = 3\). [3]

\begin{enumerate}[label=(\roman*)]
\item Write down the identity expressing $\sin 2\theta$ in terms of $\sin \theta$ and $\cos \theta$. [1]
\item Given that $\sin \alpha = \frac{1}{4}$ and $\alpha$ is acute, show that $\sin 2\alpha = \frac{1}{8}\sqrt{15}$. [3]
\item Solve, for $0° < \beta < 90°$, the equation $5 \sin 2\beta \sec \beta = 3$. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR C3  Q5 [7]}}