Year: 2011
Paper: 1
Question Number: 3
Course: LFM Pure
Section: Trigonometry 2
There were again significantly more candidates attempting this paper than last year (just over 1100), but the scores were significantly lower than last year: fewer than 2% of candidates scored above 100 marks, and the median mark was only 44, compared to 61 last year. It is not clear why this was the case. One possibility is that Questions 2 and 3, which superficially looked straightforward, turned out to be both popular and far harder than candidates anticipated. The only popular and well-answered questions were 1 and 4. The pure questions were the most popular as usual, though there was noticeable variation: questions 1–4 were the most popular, while question 7 (on differential equations) was fairly unpopular. Just over half of all candidates attempted at least one mechanics question, which one-third attempted at least one probability question, an increase on last year. The marks were surprising, though: the two best-answered questions were the pure questions 1 and 4, but the next best were statistics question 12 and mechanics question 9. The remainder of the questions were fairly similar in their marks. A number of candidates ignored the advice on the front cover and attempted more than six questions, with a fifth of candidates trying eight or more questions. A good number of those extra attempts were little more than failed starts, but still suggest that some candidates are not very effective at question-picking. This is an important skill to develop during STEP preparation. Nevertheless, the good marks and the paucity of candidates who attempted the questions in numerical order does suggest that the majority are being wise in their choices. Because of the abortive starts, I have generally restricted my attention to those attempts which counted as one of the six highest-scoring answers in the detailed comments. On occasions, candidates spent far longer on some questions than was wise. Often, this was due to an algebraic slip early on, and they then used time which could have been far better spent tackling another question. It is important to balance the desire to finish a question with an appreciation of when it is better to stop and move on. Many candidates realised that for some questions, it was possible to attempt a later part without a complete (or any) solution to an earlier part. An awareness of this could have helped some of the weaker students to gain vital marks when they were stuck; it is generally better to do more of one question than to start another question, in particular if one has already attempted six questions. It is also fine to write "continued later" at the end of a partial attempt and then to continue the answer later in the answer booklet. As usual, though, some candidates ignored explicit instructions to use the previous work, such as "Hence", or "Deduce". They will get no credit if they do not do what they are asked to! (Of course, a question which has the phrase "or otherwise" gives them the freedom to use any method of their choosing; often the "hence" will be the easiest, though.) It is wise to remember that STEP questions do require a greater facility with mathematics and algebraic manipulation than the A-level examinations, as well as a depth of understanding which goes beyond that expected in a typical sixth-form classroom. There were a number of common errors and issues which appeared across the whole paper. The first was a lack of fluency in algebraic manipulations. STEP questions often use more variables than A-level questions (which tend to be more numerical), and therefore require candidates to be comfortable engaging in extended sequences of algebraic manipulations with determination and, crucially, accuracy. This is a skill which requires plenty of practice to master. Another area of weakness is logic. A lack of confidence in this area showed up several times. In particular, a candidate cannot possibly gain full marks on a question which reads "Show that X if and only if Y" unless they provide an argument which shows that Y follows from X and vice versa. Along with this comes the need for explanations in English: a sequence of formulæ or equations with no explicit connections between them can leave the reader (and writer) confused as to the meaning. Brief connectives or explanations would help, and sometimes longer sentences are necessary. Another related issue continues to be legibility. Many candidates at some point in the paper lost marks through misreading their own writing. One frequent error was dividing by zero. On several occasions, an equation of the form xy = xz appeared, and candidates blithely divided by x to reach the conclusion y = z. Again, I give a strong reminder that it is vital to draw appropriate, clear, accurate diagrams when attempting some questions, mechanics questions in particular.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Prove the identity
\[
4\sin\theta \sin(\tfrac13\pi-\theta) \sin (\tfrac13\pi+\theta)=
\sin 3\theta\,
.
\tag{$*$}\]
\begin{questionparts}
\item By differentiating $(*)$, or otherwise, show that
\[
\cot \tfrac19\pi - \cot \tfrac29\pi + \cot \tfrac49\pi = \sqrt3\,.
\]
\item
By setting $\theta = \frac16\pi-\phi$ in $(*)$, or otherwise,
obtain a similar identity for $\cos3\theta$ and deduce that
\[
\cot \theta \cot (\tfrac13\pi-\theta) \cot (\tfrac13\pi+\theta) =\cot3\theta\,.
\]
Show that
\[
\cosec \tfrac19\pi -\cosec \tfrac59\pi +\cosec \tfrac79\pi = 2\sqrt3\,.
\]
\end{questionparts}\begin{align*}
&& LHS &= 4\sin\theta \sin(\tfrac13\pi-\theta) \sin (\tfrac13\pi+\theta) \\
&&&= 4 \sin \theta \left (\tfrac{\sqrt{3}}{2}\cos \theta - \tfrac12 \sin \theta \right)\left (\tfrac{\sqrt{3}}{2}\cos \theta + \tfrac12 \sin \theta \right) \\
&&&= 4 \sin \theta \left (\tfrac{3}{4}\cos^2 \theta - \tfrac14 \sin^2 \theta \right) \\
&&&= 3\sin \theta - 4\sin^3 \theta \\
&&&= \cos 3 \theta = RHS
\end{align*}
\begin{questionparts}
\item $\,$ \begin{align*}
&& 3 \cos 3 \theta &= \sin 3 \theta \left (\cot \theta - \cot (\tfrac13\pi - \theta) + \cot (\tfrac13\pi + \theta) \right) \\
\Rightarrow && 3 \cot 3\theta &= \cot \theta - \cot (\tfrac13\pi - \theta) + \cot (\tfrac13\pi + \theta) \\
\theta = \tfrac{\pi}{9}: && 3\cot \frac{\pi}{3} &= \cot \tfrac{\pi}{9} - \cot \tfrac{2}{9}\pi + \cot \tfrac49 \pi \\
\Rightarrow && \sqrt{3} &= \cot \tfrac{\pi}{9} - \cot \tfrac{2}{9}\pi + \cot \tfrac49 \pi
\end{align*}
\item $\,$ \begin{align*}
\theta = \tfrac16\pi - \phi && \sin(\tfrac12\pi - 3\phi) &= 4\sin(\tfrac16\pi - \phi)\sin(\phi+\tfrac16\pi)\sin(\tfrac12\pi - \phi) \\
\Rightarrow && \cos 3\phi &= 4\cos(\phi - \tfrac13\pi)\cos(\tfrac13\pi - \phi)\cos\phi \\
\Rightarrow && \cot 3\theta &= \cot \theta\cot(\phi - \tfrac13\pi)\cot(\tfrac13\pi - \phi) \tag{dividing by ($*$)} \\
\\
\frac{\d}{\d \theta}:&& -\csc^2 3\phi &= \cot3\phi \left (-\csc^2 \phi\tan \phi+\csc^2 (\tfrac13\pi - \phi) \tan (\tfrac13\pi - \phi) -\csc^2(\phi - \tfrac13\pi)\tan (\phi - \tfrac13\pi) \right) \\
\Rightarrow && \csc^23\phi\tan3\phi & = 2( \csc2\phi- \csc(\tfrac{2}{3}\pi - 2\phi)+\csc(\phi - \tfrac23\pi)) \\
\phi = \frac{1}{18}\pi: && 4\sqrt{3} &= 2(\csc \tfrac{1}{9}\pi - \csc \tfrac59\pi + \csc \tfrac79 \pi) \\
\end{align*}
and the result follows.
\end{questionparts}
This was another very popular but poorly answered question. While most candidates were able to gain some marks, few proceeded beyond the initial part of the question, giving a median mark of 5 and an upper quartile of 7. Candidates generally did well at the first part of this question, with the majority using the compound angle formulæ to expand the sines on the left hand side of the identity. A number got stuck at this point, either because they did not know sin π/3 or cos π/3 or because they made algebraic errors. About a quarter of candidates were unable to complete the proof as they did not show that sin 3θ is identical to 3 sin θ − 4 sin³θ or equivalent; some simply stated the result with no justification. On the other hand, a number of candidates offered very nice arguments using de Moivre's theorem, which was very nice to see. Some candidates used the factor formulæ, but these were in the minority. (i) Many candidates who reached this point were able to differentiate the identity correctly, which was pleasing. A significant number failed at this hurdle, though. Of those who did differentiate correctly, though, the majority failed to realise that they could divide their new identity involving cosines by the original identity to reach the stated result. Since the rest of the question depended upon this idea, barely a quarter of candidates made any further progress. There was a strong hint though: the presence of cot θ should suggest thinking about cos θ/ sin θ. (ii) Of those who had become stuck earlier, very few attempted this part, even though the first half was totally independent of part (i). Of those who did, a significant number became stuck after performing the given substitution, not realising that they could then use the identity sin(π/2 − x) = cos x. There were also a number of candidates who successfully derived the result from scratch using the compound angle formula for tan. Few candidates made it as far as the cosec equation. Of those who did, and realised that they needed to perform another division or equivalent, few were comfortable enough with the trigonometric manipulations involved to reach an expression involving cosec 2θ and thence to reach the stated result.