Year: 2021
Paper: 2
Question Number: 1
Course: LFM Pure
Section: Trigonometry 2
Candidates were generally well prepared for many of the questions on this paper, with the questions requiring more standard operations seeing the greatest levels of success. Candidates need to ensure that solutions to the questions are supported by sufficient evidence of the mathematical steps, for example when proving a given result or deducing the properties of graphs that are to be sketched. In a significant number of steps there were marks lost through simple errors such as mistakes in arithmetic or confusion of sine and cosine functions, so it is important for candidates to maintain accuracy in their solutions to these questions.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Prove, from the identities for $\cos(A \pm B)$, that
\[
\cos a \cos 3a \equiv \tfrac{1}{2}(\cos 4a + \cos 2a).
\]
Find a similar identity for $\sin a \cos 3a$.
\begin{questionparts}
\item Solve the equation
\[
4\cos x \cos 2x \cos 3x = 1
\]
for $0 \leqslant x \leqslant \pi$.
\item Prove that if
\[
\tan x = \tan 2x \tan 3x \tan 4x \qquad (\dagger)
\]
then $\cos 6x = \tfrac{1}{2}$ or $\sin 4x = 0$.
Hence determine the solutions of equation $(\dagger)$ with $0 \leqslant x \leqslant \pi$.
\end{questionparts}\begin{align*}
&& \cos(A \pm B) &= \cos A \cos B \mp \sin A \sin B \\
A = a, B = 3a&& \cos 4a + \cos 2a &= 2\cos 3a \cos a \\
\Rightarrow && \cos a \cos 3a &= \tfrac12(\cos 4a + \cos 2a) \\
\\
&& \sin(A \pm B) &= \sin A \cos B \pm \cos A \sin B \\
&& \sin 4a + \sin(- 2a) &= 2 \sin a \cos 3a \\
\Rightarrow && \sin a \cos 3a &= \tfrac12 (\sin 4a - \sin 2a)
\end{align*}
\begin{questionparts}
\item $\,$ \begin{align*}
&& 1 &= 4 \cos x \cos 2x \cos 3x \\
&&&= 2(\cos 4x +\cos 2x)\cos 2x \\
c = \cos 2x:&&&= 2(2c^2-1+c)c \\
\Rightarrow && 0 &= 4c^3+2c^2-2c-1 \\
&&&= (2c+1)(2c^2-1) \\
\Rightarrow && \cos 2x &= -\frac12 \\
\Rightarrow && x &= \frac{\pi}{3}, \frac{2\pi}{3} \\
&& \cos 2x &= \pm \frac1{\sqrt2} \\
\Rightarrow && x&= \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8}
\end{align*}
\item $\,$ \begin{align*}
&& \tan x &= \tan 2x \tan 3x \tan 4x \\
\Rightarrow &&1 &= \frac{\cos x\sin 2x \sin 3x \sin 4x}{\sin x \cos 2x \cos 3x \cos 4x} \\
&&&= \frac{\sin 2x \sin 4x (\sin4 x + \sin 2x)}{\cos 2x \cos 4x (\sin 4x - \sin 2x)} \\
&&&= \frac{(\cos 2x - \cos 6x) (\sin4 x + \sin 2x)}{(\cos 6x + \cos 2x) (\sin 4x - \sin 2x)} \\
\Rightarrow && 0 &= 2\cos 6x \sin 4x - 2\cos 2x \sin 2 x\\
&&&= \sin 4 x (2 \cos 6x - 1) \\
\Rightarrow && \sin 4x &= 0 \\
\text{ or }&& \cos 6x &= \frac12
\end{align*}
$\sin 4x = 0 \Rightarrow x = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi$
$\cos 6x = \frac12 \Rightarrow x = \frac{\pi}{18}, \frac{5\pi}{18},\frac{7\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}$.
We should verify these work, since not all of them will, especially where $\sin 4x = 0$, so our final answer is
$x = 0, \pi, \frac{\pi}{18}, \frac{5\pi}{18},\frac{7\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}$
\end{questionparts}
Many candidates were able to prove the given identity in the opening sentence of the question, although there were a large number of attempts that took the approach of expressing both the left and right sides of the identity in terms of cos a and sin a. Those who recognised that the result followed quickly from applying the identity for cos(A ± B) for A = 3a and B = a were then much more able to find the similar identity for sin a cos 3a. Many candidates were able to apply the identity from the start of the question to the equation in part (i) and went on to solve the equation successfully in the required interval. A small number of candidates did not realise that it was not necessary to express the equation as a polynomial in cos a and so encountered a more difficult polynomial to solve in order to reach the solutions. In many cases this did not result in the correct set of solutions being found. Part (ii) was less well attempted in general. While almost all candidates realised that writing the tangent functions in terms of sine and cosine would be useful many were not able to rearrange into a sufficiently useful form to make further progress on the question. Those who did often managed to reach the required result without too much difficulty. Candidates had little problem finding the full set of solutions to the two equations deduced in the first section of (ii), but then most failed to realise that some of those solutions were not possible as the equation involved tangent functions.