Year: 2009
Paper: 1
Question Number: 4
Course: LFM Pure
Section: Introduction to trig
There were significantly more candidates attempting this paper again this year (over 900 in total), and the scores were pleasing: fewer than 5% of candidates failed to get at least 20 marks, and the median mark was 48. The pure questions were the most popular as usual; about two-thirds of candidates attempted each of the pure questions, with the exceptions of question 2 (attempted by about 90%) and question 5 (attempted by about one third). The mechanics questions were only marginally more popular than the probability and statistics questions this year; about one quarter of the candidates attempted each of the mechanics questions, while the statistics questions were attempted by about one fifth of the candidates. A significant number of candidates ignored the advice on the front cover and attempted more than six questions. In general, those candidates who submitted answers to eight or more questions did fairly poorly; very few people who tackled nine or more questions gained more than 60 marks overall (as only the best six questions are taken for the final mark). This suggests that a skill lacking in many students attempting STEP is the ability to pick questions effectively. This is not required for A-levels, so must become an important part of STEP preparation. Another "rubric"-type error was failing to follow the instructions in the question. In particular, when a question says "Hence", the candidate must make (significant) use of the preceding result(s) in their answer if they wish to gain any credit. In some questions (such as question 2), many candidates gained no marks for the final part (which was worth 10 marks) as they simply quoted an answer without using any of their earlier work. There were a number of common errors which appeared across the whole paper. These included a noticeable weakness in algebraic manipulations, sometimes indicating a serious lack of understanding of the mathematics involved. As examples, one candidate tried to use the misremembered identity cos β = sin √(1 − β²), while numerous candidates made deductions of the form "if a² + b² = c², then a + b = c" at some point in their work. Fraction manipulations are also notorious in the school classroom; the effects of this weakness were felt here, too. Another common problem was a lack of direction; writing a whole page of algebraic manipulations with no sense of purpose was unlikely to either reach the requested answer or gain the candidate any marks. It is a good idea when faced with a STEP question to ask oneself, "What is the point of this (part of the) question?" or "Why has this (part of the) question been asked?" Thinking about this can be a helpful guide. One aspect of this is evidenced by pages of formulæ and equations with no explanation. It is very good practice to explain why you are doing the calculation you are, and to write sentences in English to achieve this. It also forces one to focus on the purpose of the calculations, and may help avoid some dead ends. Finally, there is a tendency among some candidates when short of time to write what they would do at this point, rather than using the limited time to actually try doing it. Such comments gain no credit; marks are only awarded for making progress in a question. 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.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
The sides of a triangle have lengths
$p-q$, $p$ and $p+q$, where $p>q> 0\,$.
The largest
and smallest angles of the triangle are $\alpha$ and $\beta$,
respectively.
Show by means of the cosine rule that
\[ 4(1-\cos\alpha)(1-\cos\beta) = \cos\alpha + \cos\beta
\,.
\]
In the case $\alpha = 2\beta$,
show that $\cos\beta=\frac34$ and hence find the
ratio of the lengths of the sides of the triangle.The largest angle will be opposite the side with length $p+q$. Similarly the smallest angle will be opposite the side with length $p-q$.
The cosine rule tells us that:
\begin{align*}
&& (p+q)^2 &= p^2 + (p-q)^2 - 2p(p-q) \cos \alpha \\
&& 0 &= p(p-4q-2(p-q)\cos \alpha)\\
&& 0 &= p(1-2\cos \alpha) + q(2\cos \alpha - 4)\\
\Rightarrow && \frac{p}{q} & = \frac{4-2 \cos \alpha}{1-2 \cos \alpha} \\
&& (p-q)^2 &= p^2 + (p+q)^2 - 2p(p+q) \cos \beta \\
&& 0 &= p(p+4q-2(p+q) \cos \beta) \\
&& 0 &= p(1-2\cos \beta)+q(4-2\cos \beta) \\
\Rightarrow && \frac{p}{q} &= \frac{2\cos \beta - 4}{1-2\cos \beta} \\
\Rightarrow && \frac{4-2 \cos \alpha}{1-2 \cos \alpha} &= \frac{2\cos \beta - 4}{1-2\cos \beta} \\
\Rightarrow && (2-\cos \alpha)(1-2\cos \beta) &= (\cos \beta - 2)(1 - 2 \cos \alpha) \\
\Rightarrow && 2 - \cos \alpha -4\cos \beta+2\cos \alpha \cos \beta &= \cos \beta - 2-2\cos \alpha \cos \beta + 4 \cos \alpha \\
\Rightarrow && 4-4\cos \alpha - 4\cos \beta+4\cos \alpha\cos \beta &= \cos \alpha + \cos \beta \\
\Rightarrow && 4(1-\cos \alpha)(1-\cos \beta) &= \cos \alpha + \cos \beta
\end{align*}
If $\alpha = 2 \beta$, and let $c = \cos \beta$
\begin{align*}
&& 4 (1- \cos 2 \beta)(1-\cos \beta) &= \cos 2 \beta + \cos \beta \\
\Rightarrow && 4(1-(2c^2-1))(1-c) &= 2c^2-1+c\\
\Rightarrow && 8(1+c)(1-c)^2 &= (2c-1)(c+1) \\
\Rightarrow && 0 &= (c+1)(8(1-c)^2-(2c-1)) \\
&&&= (c+1)(8c^2-18c+9) \\
&&&= (c+1)(4c-3)(2c-3) \\
\end{align*}
Therefore $c = -1, \frac32, \frac34$. Clearly $\cos \beta \neq -1, \frac32$, since they are not valid angles in a triangle (or valid values of $\cos \beta$).
$\frac{p}{q} = \frac{2 \cdot \frac34-4 }{1 - 2\cdot \frac34} = \frac{3-8}{2-3} = 5$ so $4:5:6$
This question was found to have a moderate level of difficulty. There were some very good attempts as well as some very poor ones. The initial part of the question confused some candidates; the sketches produced showed that the statement: "The largest and smallest angles of the triangle are α and β, respectively" caused a surprising amount of difficulty. The use of "respectively", in particular, is standard mathematical language and should be recognised—a number of candidates swapped the two angles. (It might be that some genuinely thought that the angles were the other way round, but most simply probably misunderstood the question.) There were also a fair number of candidates who labelled the angle opposite p as either α or β, which prevented them from making much further progress in the question. Most candidates were able to correctly use the cosine rule in the context of this question, which was very pleasing. Manipulating it to deduce the required equation was much harder, though: as usual, the lack of algebraic fluency was the sticking point. There were many careless errors, including sign errors, transcription errors and the like. Also, a number of candidates did not simplify the algebraic fractions when they could, thereby making their task even more challenging. Many demonstrated that they did not understand how to add fractions, and the monstrous messes which resulted were quite painful to see. The straightforward method for approaching this question was to substitute expressions for cos α and cos β into the left and right hand sides of the required equation and to show that they turned out to be equal. A significant number of candidates did not realise that this would be useful, and instead produced pages of algebra leading nowhere. In the second part of the question, a number of candidates who had not succeeded in the first part nonetheless realised that they could use the given result to solve the second part, and some were successful at this. There were some good attempts at the second part. The sticking points were once again mostly algebraic: some did not know the double angle formulæ or they used incorrect expressions for them; of those who did know them, many then erred in their expansion of the brackets. The next sticking point was solving the cubic; some did not attempt to do so, while others made algebraic slips. Those who did succeed in deducing cos β usually went on to correctly find the side lengths and hence the required ratio.