| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2007 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Prove identity with double/compound angles |
| Difficulty | Challenging +1.2 This is a structured Further Maths question with clear guidance at each step. Part (a) is algebraic manipulation following a given substitution (routine for FP2 students). Part (b) applies the telescoping series method explicitly signposted in the question. While it requires careful bookkeeping with the summation, the technique is standard and the question provides substantial scaffolding, making it moderately above average difficulty. |
| Spec | 1.05l Double angle formulae: and compound angle formulae4.06b Method of differences: telescoping series |
7
\begin{enumerate}[label=(\alph*)]
\item Use the identity $\tan ( A - B ) = \frac { \tan A - \tan B } { 1 + \tan A \tan B }$ with $A = ( r + 1 ) x$ and $B = r x$ to show that
$$\tan r x \tan ( r + 1 ) x = \frac { \tan ( r + 1 ) x } { \tan x } - \frac { \tan r x } { \tan x } - 1$$
(4 marks)
\item Use the method of differences to show that
$$\tan \frac { \pi } { 50 } \tan \frac { 2 \pi } { 50 } + \tan \frac { 2 \pi } { 50 } \tan \frac { 3 \pi } { 50 } + \ldots + \tan \frac { 19 \pi } { 50 } \tan \frac { 20 \pi } { 50 } = \frac { \tan \frac { 2 \pi } { 5 } } { \tan \frac { \pi } { 50 } } - 20$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2007 Q7 [9]}}