Year: 2019
Paper: 2
Question Number: 4
Course: LFM Pure
Section: Trigonometry 2
The Pure questions were again the most popular of the paper, with only one of the questions attempted by fewer than half of the candidates (of the remaining four questions, only question 9 was attempted by more than a quarter of the candidates). In many of the questions candidates were often unable to make good use of the results shown in the earlier parts of the question in order to solve the more complex later parts. Nevertheless, some good solutions were seen to all of the questions. For many of the questions, solutions were seen in which the results were reached, but without sufficient justification of some of the steps.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
You are not required to consider issues of convergence in this question.
For any sequence of numbers $a_1, a_2, \ldots, a_m, \ldots, a_n$, the notation $\prod_{i=m}^{n} a_i$ denotes the product $a_m a_{m+1} \cdots a_n$.
\begin{questionparts}
\item Use the identity $2 \cos x \sin x = \sin(2x)$ to evaluate the product $\cos(\frac{\pi}{9}) \cos(\frac{2\pi}{9}) \cos(\frac{4\pi}{9})$.
\item Simplify the expression
$$\prod_{k=0}^{n} \cos\left(\frac{x}{2^k}\right) \quad (0 < x < \frac{1}{2}\pi).$$
Using differentiation, or otherwise, show that, for $0 < x < \frac{1}{2}\pi$,
$$\sum_{k=0}^{n} \frac{1}{2^k} \tan\left(\frac{x}{2^k}\right) = \frac{1}{2^n} \cot\left(\frac{x}{2^n}\right) - 2 \cot(2x).$$
\item Using the results $\lim_{\theta\to 0} \frac{\sin \theta}{\theta} = 1$ and $\lim_{\theta\to 0} \frac{\tan \theta}{\theta} = 1$, show that
$$\prod_{k=1}^{\infty} \cos\left(\frac{x}{2^k}\right) = \frac{\sin x}{x}$$
and evaluate
$$\sum_{j=2}^{\infty} \frac{1}{2^{j-2}} \tan\left(\frac{\pi}{2^j}\right).$$
\end{questionparts}\begin{questionparts}
\item
\begin{align*}\cos(\frac{\pi}{9}) \cos(\frac{2\pi}{9}) \cos(\frac{4\pi}{9}) &= \frac{\sin(\frac{2\pi}{9}) \cos(\frac{2\pi}{9}) \cos(\frac{4\pi}{9})}{2 \sin \frac{\pi}{9}} \\
&= \frac{\sin(\frac{4\pi}{9})\cos(\frac{4\pi}{9})}{4 \sin \frac{\pi}{9}} \\
&= \frac{\sin(\frac{8\pi}{9})}{8 \sin \frac{\pi}{9}} \\
&= \frac{1}{8}
\end{align*}
\item Let $\displaystyle P_n = \prod_{k=0}^{n} \cos\left(\frac{x}{2^k}\right)$.
Claim: $P_n = \frac{\sin 2x}{2^{n+1} \sin \l \frac{x}{2^n} \r}$.
Proof: This is true for $n = 0$, assume true for $n-1$
\begin{align*}
\sin\l \frac{x}{2^{n}} \r P_n &= P_{n-1} \cos\l \frac{x}{2^{n}} \r \sin\l \frac{x}{2^{n}} \r \\
&= P_{n-1} \frac{1}{2} \sin\l \frac{x}{2^{n-1}} \r \\
&= \frac{\sin 2x}{2^{n} \sin \l \frac{x}{2^{n-1}}\r} \frac{1}{2} \sin\l \frac{x}{2^{n}} \r \\
&= \frac{\sin 2x}{2^{n+1}}
\end{align*}
Hence $P_n = \frac{\sin 2x}{2^{n+1} \sin \l \frac{x}{2^n}\r}$
Taking logs, we determine that:
\begin{align*}
&& \sum_{k=0}^n \ln \cos \l \frac{x}{2^k} \r &= \ln \sin 2x - \ln \sin \l \frac{x}{2^n} \r - (n+1) \ln 2 \\
\Rightarrow && \sum_{k=0}^n \frac{1}{2^k} \tan \l \frac{x}{2^k} \r &= -2 \cot 2x + \frac{1}{2^n} \cot \l \frac{x}{2^n} \r - 0 \\
\end{align*}
as required.
\item
As $n \to \infty$ $\frac{x}{2^n} \to 0$, so $\frac{\sin \frac{x}{2^n}}{\frac{x}{2^n}} = \frac{2^n \sin \frac{x}{2^n}}{x} \to 1$
\begin{align*}\prod_{k=1}^{\infty} \cos\left(\frac{x}{2^k}\right) &= \lim_{n \to \infty} \frac{\sin x}{2^n \sin \l \frac{x}{2^n} \r} \\
&= \lim_{n \to \infty} \frac{\sin x}{x \frac{2^n \sin \l \frac{x}{2^n} \r}{x} } \\
&= \lim_{n \to \infty} \frac{\sin x}{x} \\
\end{align*}
\begin{align*}
\sum_{j=2}^{\infty} \frac{1}{2^{j-2}} \tan\left(\frac{\pi}{2^j}\right) &= \sum_{j=0}^{\infty} \frac{1}{2^{j}} \tan\left(\frac{1}{2^j}\frac{\pi}{4}\right) \\
&= \lim_{n \to \infty} \l -2 \cot \frac{\pi}{2} + \frac{1}{2^n} \cot \l \frac{\pi}{4 \cdot 2^n} \r\r \\
&= \frac{4}{\pi} \lim_{n \to \infty} \l \frac{1}{2^n} \frac{\pi}{4} \cot \l \frac{\pi}{4 \cdot 2^n} \r\r \\
&\to \frac{\pi}{4}
\end{align*}
\end{questionparts}
This was a well-answered question, but also one in which a fairly large number of solutions scored very low marks. The majority of candidates were able to evaluate the first product using the identity provided and most were then able to apply the same technique to simplify the first expression in part (i). Many students then differentiated, but some then struggled to manage the notation correctly to reach the second result requested in part (ii). Part (iii) required some care to ensure that the sums and products were over the correct range, but those who managed to adjust correctly for this were then able to reach the required results.