OCR H240/01 2018 December — Question 12 9 marks

Exam BoardOCR
ModuleH240/01 (Pure Mathematics)
Year2018
SessionDecember
Marks9
TopicAddition & Double Angle Formulae
TypeDerive triple angle then solve equation
DifficultyChallenging +1.2 Part (a) is a guided derivation using standard addition formulae—straightforward but requires careful algebra with multiple steps. Part (b) requires rearranging to form a cubic in tan θ, then using the triple angle result to argue about the number of solutions graphically or algebraically, which demands more mathematical maturity and insight than typical A-level questions but is still within reach with the scaffolding provided.
Spec1.01a Proof: structure of mathematical proof and logical steps1.05l Double angle formulae: and compound angle formulae

12
  1. By first writing \(\tan 3 \theta\) as \(\tan ( 2 \theta + \theta )\), show that \(\tan 3 \theta = \frac { 3 \tan \theta - \tan ^ { 3 } \theta } { 1 - 3 \tan ^ { 2 } \theta }\).
  2. Hence show that there are always exactly two different values of \(\theta\) between \(0 ^ { \circ }\) and \(180 ^ { \circ }\) which satisfy the equation \(3 \tan 3 \theta = \tan \theta + k\),
    where \(k\) is a non-zero constant. \section*{END OF QUESTION PAPER} \section*{OCR
    Oxford Cambridge and RSA}

12
\begin{enumerate}[label=(\alph*)]
\item By first writing $\tan 3 \theta$ as $\tan ( 2 \theta + \theta )$, show that $\tan 3 \theta = \frac { 3 \tan \theta - \tan ^ { 3 } \theta } { 1 - 3 \tan ^ { 2 } \theta }$.
\item Hence show that there are always exactly two different values of $\theta$ between $0 ^ { \circ }$ and $180 ^ { \circ }$ which satisfy the equation\\
$3 \tan 3 \theta = \tan \theta + k$,\\
where $k$ is a non-zero constant.

\section*{END OF QUESTION PAPER}
\section*{OCR \\
 Oxford Cambridge and RSA}
\end{enumerate}

\hfill \mbox{\textit{OCR H240/01 2018 Q12 [9]}}