Year: 2015
Paper: 2
Question Number: 5
Course: LFM Pure
Section: Trigonometry 2
As in previous years the Pure questions were the most popular of the paper with questions 1, 2 and 6 the most popular. The least popular questions on the paper were questions 8, 11 and 13 with fewer than 250 attempts for each of them. There were many examples of solutions in this paper that were insufficiently well explained given that the answer to be reached had been provided in the question.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1484.9
Banger Comparisons: 1
In this question, the $\mathrm{arctan}$ function satisfies $0\le \arctan x <\frac12 \pi$ for $x\ge0\,$.
\begin{questionparts}
\item Let
\[
S_n= \sum_{m=1}^n \arctan \left(\frac1 {2m^2}\right)
\,,
\]
for $n=1, 2, 3, \ldots$ . Prove by induction that
\[
\tan S_n = \frac n {n+1} \,.
\]
Prove also that
\[
S_n = \arctan \frac n {n+1} \,.
\]
\item In a triangle $ABC$, the lengths of the sides $AB$ and $BC$ are $4n^2$ and $4n^4-1$, respectively, and the angle at $B$ is a right angle. Let $\angle BCA = 2\alpha_n$. Show that
\[
\sum_{n=1}^\infty \alpha_n = \tfrac14\pi \,.
\]
\end{questionparts}\begin{questionparts}
\item Claim: $\tan S_n = \frac n {n+1}$
Proof: (By Induction)
Base case: ($n=1$):
\begin{align*}
&& \tan \left ( \sum_{m=1}^1 \arctan \left ( \frac{1}{2m^2} \right) \right) &= \tan \left ( \arctan \left ( \frac{1}{2} \right) \right) \\
&&&= \frac12 = \frac{1}{1+1}
\end{align*}
Therefore the base case is true.
Inductive step: Suppose our statement is true for some $n = k$, ie
\begin{align*}
&& \frac{k}{k+1} &= \tan \left ( \sum_{m=1}^k \arctan \left ( \frac{1}{2m^2} \right) \right) \\
\Rightarrow && \tan S_{k+1} &= \tan \left ( \sum_{m=1}^k \arctan \left ( \frac{1}{2m^2} \right) + \arctan \left ( \frac{1}{2 (k+1)^2} \right) \right) \\
&&&= \frac{\tan S_k + \tan \left ( \arctan \left ( \frac{1}{2 (k+1)^2} \right) \right)}{1-\tan S_k \tan \left ( \arctan \left ( \frac{1}{2 (k+1)^2} \right) \right)} \\
&&&= \frac{\frac{k}{k+1} + \frac{1}{2(k+1)^2}}{1-\frac{k}{k+1} \frac{1}{2(k+1)^2}} \\
&&&= \frac{2k(k+1)^2+(k+1)}{2(k+1)^3-k} \\
&&&= \frac{k+1}{(k+1)+1}
\end{align*}
Therefore it is true for $n=k+1$.
Conclusion: Therefore by the principle of mathematical induction since our statement is true for $n=1$ and if it is true for $n=k$ it is true for $n=k+1$ it is true for all $n\geq1$
Since $S_n < \frac12 \pi$ for all $n$, we must have $\arctan \frac{n}{n+1} = S_n$
\item $\tan (2\alpha_n) = \frac{4n^2}{4n^4-1} = \frac{2n^2+2n^2}{(2n^2)(2n^2)-1} = \frac{\frac{1}{2n^2}+\frac{1}{2n^2}}{1-\frac{1}{2n^2}\frac{1}{2n^2}} \Rightarrow \tan (\alpha_n) = \arctan \frac{1}{2n^2}$. In particular $\displaystyle \sum_{n=1}^{N} \alpha_n = \arctan \frac{n}{n+1} \Rightarrow \sum_{n=1}^{\infty} \alpha_n \to \arctan 1 = \frac{\pi}{4} $
\end{questionparts}
This question had one of the lower average marks for the Pure maths questions on the paper. Most candidates were able to produce a proof by induction for the first part, but the vast majority failed to realise that there was more that needed to be done to prove the result stated in terms of arctan. As is the case for a number of other questions, candidates need to give a clear explanation of each step of the solution. Where candidates identified the relationship between the two parts of the question the second part was generally well attempted.