Year: 2018
Paper: 2
Question Number: 4
Course: LFM Pure
Section: Trigonometry 2
The pure questions were again the most popular of the paper, with only two of those questions being attempted by fewer than half of the candidates (none of the other questions was attempted by more than half of the candidates). Good responses were seen to all of the questions, but in many cases, explanations lacked sufficient detail to be awarded full marks. Candidates should ensure that they are demonstrating that the results that they are attempting to apply are valid in the cases being considered. In several of the questions, later parts involve finding solutions to situations that are similar to earlier parts of the question. In general candidates struggled to recognise these similarities and therefore spent a lot of time repeating work that had already been done, rather than simply observing what the result must be.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
In this question, you may use the following identity without proof:
\[ \cos A + \cos B = 2\cos\tfrac12(A+B) \, \cos \tfrac12(A-B) \;. \]
\begin{questionparts}
\item Given that $0\le x \le 2\pi$, find all the values of $x$ that satisfy the equation
\[
\cos x + 3\cos 2x + 3\cos 3 x + \cos 4x= 0
\,.
\]
\item Given that $0\le x \le \pi$ and $0\le y \le \pi$ and that
\[
\cos (x+y) + \cos (x-y) -\cos2x = 1
\,,
\]
show that either $x=y$ or $x$ takes one specific value which you should find.
\item Given that $0\le x \le \pi$ and $0\le y \le \pi\,$, find the values of $x$ and $y$ that satisfy the equation
\[
\cos x + \cos y -\cos (x+y) = \tfrac32
\,.
\]
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& 0 &= \cos x + 3 \cos 2x + 3 \cos 3x + \cos 4 x \\
&&&= \cos x + \cos 4x + 3 \left (\cos 2x + \cos 3 x \right) \\
&&&= 2 \cos \frac{5x}{2} \cos \frac{3x}{2} + 6 \cos \frac{x}{2}\cos\frac{5x}{2} \\
&&&= 2 \cos \frac{5x}{2} \left (\cos \frac{3x}{2} + 3 \cos \frac{x}{2} \right)\\
&&&= 2 \cos \frac{5x}{2} \left ( \cos \frac{2x}{2}\cos \frac{x}{2} - \sin \frac{2x}{2} \sin \frac{x}{2}+3 \cos \frac{x}{2} \right) \\
&&&= 2 \cos \frac{5x}{2} \left ( \left (2\cos^2 \frac{x}{2} - 1 \right)\cos \frac{x}{2} - 2\sin \frac{x}{2} \cos \frac{x}{2} \sin \frac{x}{2}+3 \cos \frac{x}{2} \right) \\
&&&= 2 \cos \frac{5x}{2} \left ( 4\cos^3 \frac{x}{2} \right) \\
&&&= 8 \cos \frac{5x}{2} \cos^3 \frac{x}{2} \\
\Rightarrow && \frac{x}{2} &= \frac{\pi}{2}, \frac{3\pi}{2}, \cdots \\
&& \frac{5x}{2} &= \frac{\pi}{2}, \frac{3\pi}{2}, \cdots \\
\Rightarrow && x &= \frac{\pi}{5}, \frac{3\pi}{5}, \pi, \frac{7\pi}{5}, \frac{9\pi}{5}
\end{align*}
\item $\,$ \begin{align*}
&& 1 &= \cos (x + y) + \cos(x-y) - \cos 2x \\
&&&= 2 \cos x \cos y - 2\cos^2 x + 1 \\
\Rightarrow && 0 &= \cos x (\cos y - \cos x) \\
\Rightarrow && 0 &=\cos x \left ( \cos y + \cos (\pi - x) \right) \\
&&&= 2\cos x \cos \frac{y+x-\pi}{2} \cos \frac{y-x+\pi}{2} \\
\Rightarrow && x &= \frac{\pi}{2} \\
&& y+x - \pi&= \pi ,3\pi, \cdots \\
&& y-x + \pi&=\pi, 3 \pi, \cdots \\
\Rightarrow && x &= \frac{\pi}{2} \\
&& y+x &= 2\pi \Rightarrow x = y = \pi \\
&& y&= x
\end{align*}
So the only solutions are $x =y$ and $x = \frac{\pi}{2}$
\item $\,$ \begin{align*}
&& \frac32 &= \cos x + \cos y - \cos (x+y) \\
&&&= 2 \cos \frac{x+y}{2} \cos \frac{x-y}{2} - 2 \cos^2 \frac{x+y}{2} + 1 \\
\Rightarrow && \frac14 &= \cos \frac{x+y}{2} \left ( \cos \frac{x-y}{2} - \cos \frac{x+y}{2} \right) \\
\Rightarrow && 0 &= \cos^2 \frac{x+y}{2} - \cos \frac{x-y}{2}\cos \frac{x+y}{2} + \frac14 \\
\Rightarrow && \cos \frac{x+y}{2} &= \frac{\cos \frac{x-y}{2} \pm \sqrt{\cos^2 \frac{x-y}{2}-1}}{2} \\
\Rightarrow && \cos \frac{x-y}{2} &= \pm 1\\
&& \cos \frac{x+y}{2} &= \pm \frac12 \\
\Rightarrow && x-y &= -4\pi, 0, 4\pi, \cdots \\
\Rightarrow && x &= y \\
\Rightarrow && \cos x &= \frac12 \\
\Rightarrow && x &= \frac{\pi}{3} \\
\Rightarrow && (x, y) &= \left ( \frac{\pi}{3}, \frac{\pi}{3}\right)
\end{align*}
\end{questionparts}
This was the second most popular question on the paper. The average mark scored by candidates attempting this question was the highest of all the pure questions. The first two parts of the question were generally well done with candidates showing confidence in applying the given identity and then factorising and solving the resulting equation. A sizeable minority struggled to deal with the given range for and so were unable to find all the solutions. In part (ii) many candidates were able to explain clearly why cos = cos leads to = . Part (iii), however, proved to be difficult for the majority of candidates. A small number of candidates did manage to find the quadratic equation and most of these were able to proceed and complete the question fully. Most candidates did not score very many marks in this section.