Year: 2011
Paper: 2
Question Number: 4
Course: LFM Pure
Section: Trigonometry 2
There were just under 1000 entries for paper II this year, almost exactly the same number as last year. After the relatively easy time candidates experienced on last year's paper, this year's questions had been toughened up significantly, with particular attention made to ensure that candidates had to be prepared to invest more thought at the start of each question – last year saw far too many attempts from the weaker brethren at little more than the first part of up to ten questions, when the idea is that they should devote 25-40 minutes on four to six complete questions in order to present work of a substantial nature. It was also the intention to toughen up the final "quarter" of questions, so that a complete, or nearly-complete, conclusion to any question represented a significant (and, hopefully, satisfying) mathematical achievement. Although such matters are always best assessed with the benefit of hindsight, our efforts in these areas seem to have proved entirely successful, with the vast majority of candidates concentrating their efforts on four to six questions, as planned. Moreover, marks really did have to be earned: only around 20 candidates managed to gain or exceed a score of 100, and only a third of the entry managed to hit the half-way mark of 60. As in previous years, the pure maths questions provided the bulk of candidates' work, with relatively few efforts to be found at the applied ones. Questions 1 and 2 were attempted by almost all candidates; 3 and 4 by around three-quarters of them; 6, 7 and 9 by around half; the remaining questions were less popular, and some received almost no "hits". Overall, the highest scoring questions (averaging over half-marks) were 1, 2 and 9, along with 13 (very few attempts, but those who braved it scored very well). This at least is indicative that candidates are being careful in exercising some degree of thought when choosing (at least the first four) 'good' questions for themselves, although finding six successful questions then turned out to be a key discriminating factor of candidates' abilities from the examining team's perspective. Each of questions 4-8, 11 & 12 were rather poorly scored on, with average scores of only 5.5 to 6.6.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1470.8
Banger Comparisons: 2
\begin{questionparts}
\item Find all the values of $\theta$, in the range $0^\circ < \theta < 180^\circ$, for which $\cos\theta=\sin 4\theta$. Hence show that
\[
\sin 18^\circ = \frac14\left( \sqrt 5 -1\right).
\]
\item Given that
\[
4\sin^2 x + 1 = 4\sin^2 2x
\,,
\]
find all possible values of $\sin x\,$, giving your answers in the form $p+q\sqrt5$ where $p$ and $q$ are rational numbers.
\item Hence find two values of $\alpha$ with $0^\circ < \alpha < 90^\circ$ for which
\[
\sin^23\alpha + \sin^25\alpha = \sin^2 6\alpha\,.
\]
\end{questionparts}\begin{questionparts}
\item Note that $\cos \theta = \sin (90^\circ - \theta)$ so
\begin{align*}
&& \sin(90^\circ - \theta) &= \sin 4 \theta\\
&& 90^\circ - \theta &= 4\theta +360^{\circ}k \\
&& 90^\circ + \theta &= 4\theta +360^{\circ}k \\
\Rightarrow && 5\theta &= 90^\circ, 450^\circ, 810^\circ, \cdots \\
&& 3 \theta &= 90^\circ, 450^\circ, \cdots \\
\Rightarrow && \theta &= 18^\circ, 90^\circ, 162^\circ, \ldots \\
&& \theta &= 30^\circ, 150^\circ, \ldots
\end{align*}
Therefore $\theta = 8^\circ, 30^\circ, 90^\circ, 150^\circ, 162^\circ$.
Note also that:
\begin{align*}
&& 0 &= \sin 4 \theta - \cos \theta \\
&&&= 2 \sin 2 \theta \cos 2 \theta- \cos \theta \\
&&&= 4 \sin \theta \cos \theta \cos 2 \theta - \cos \theta \\
&&&= \cos \theta \left (4 \sin \theta (1- 2\sin^2 \theta) - 1 \right) \\
&&&= \cos \theta \left (-8\sin^3 \theta +4\sin \theta - 1 \right) \\
&&&= \cos \theta (1 - 2 \sin \theta)(4 \sin^2 \theta+2\sin \theta -1)\\
\cos \theta = 0: && \theta &= 90^\circ \\
\sin \theta = \frac12: && \theta &= 30^{\circ} \\
&& \theta &= \sin^{-1} \left ( \frac{-1\pm \sqrt5}{4} \right)
\end{align*}
Therefore $\sin 18^{\circ} = \frac{\pm \sqrt{5}-1}{4}$, but since $\sin 18^{\circ} > 0$ it must be the positive version.
\item $\,$ \begin{align*}
&& 4 \sin^2 x + 1 &= 4 \sin^2 2 x \\
&&&= 16 \sin^2 x \cos^2 x \\
&&&= 16 \sin^2 x (1- \sin^2 x) \\
\Rightarrow && 0 &= 16y^2 -12y+1 \\
\Rightarrow && \sin^2 x &= \frac{3\pm \sqrt5}{8} \\
&&&= \left ( \frac{1 \pm \sqrt5}{4} \right)^2 \\
\Rightarrow && \sin x &= \pm \frac{1 \pm \sqrt{5}}{4}
\end{align*}
\item $\,$ \begin{align*}
&& \sin^2 x + \frac1{2^2} &= \sin^2 2x
\end{align*}
So if we can have $\sin 5x = \pm \frac12$ and $\sin 3x = \pm \frac{1 \pm \sqrt5}{4}$ then we are good, ie
\begin{align*}
&& 5x &= 30^{\circ}, 150^{\circ}, 210^{\circ}, 330^{\circ}, 390^{\circ}, \cdots \\
\Rightarrow && x &= 6^{\circ}, 30^{\circ}, 42^{\circ}, 66^{\circ}, 78^{\circ} \\
\Rightarrow && 3x &= \boxed{18^{\circ}}, \cancel{90^{\circ}}, \boxed{126^{\circ}}, \boxed{198^{\circ}}, \boxed{78^{\circ}}
\end{align*}
So our solutions are $x = 6^{\circ}, 42^{\circ}, 66^{\circ}, 78^{\circ}$ although it's interesting to note that $x = 45^{\circ}$ is another solution
\end{questionparts}
This question was the first of the really popular ones to attract relatively low scores overall. In the opening part, it had been expected that candidates would employ that most basic of trig. identities, sinA = cos(90° – A), in order to find the required values of θ, but the vast majority went straight into double-angles and quadratics in terms of sinθ instead, which had been expected to follow the initial work; this meant that many candidates were unable to explain convincingly why the given value of sin18° was as claimed. Despite the relatively straightforward trig. methods that were required in this question, with part (ii) broadly approachable in the same way as the second part of (i), the lack of a clear-minded strategy proved to be a big problem for most attempters, and the connection between parts (ii) and (iii) was seldom spotted – namely, to divide through by 4 and realise that sin5θ must be ±½. Many spotted the solution θ = 6°, but few got further than this because they were stuck exclusively on sin30° = ½.