Year: 2010
Paper: 1
Question Number: 3
Course: LFM Pure
Section: Trigonometry 2
There were significantly more candidates attempting this paper than last year (just over 1000), and the scores were much higher than last year (presumably due to the easier first question): fewer than 2% of candidates scored less than 20 marks overall, and the median mark was 61. The pure questions were the most popular as usual, though there was much more variation than in some previous years: questions 1, 3, 4 and 6 were the most popular, while question 7 (on vectors) was intensely unpopular. About half of all candidates attempted at least one mechanics question, and 15% attempted at least one probability question. The marks were unsurprising: the pure questions generally gained the better marks, while the mechanics and probability questions generally had poorer marks. A sizeable 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 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 often restricted my attention below to those attempts which counted as one of the six highest-scoring answers, and referred to these as "significant attempts". The majority of candidates did begin with question 1 (presumably as it appeared to be the easiest), but some spent far longer on it than was wise. Some attempts ran to over eight pages in length, especially when they had made an algebraic slip early on, and used time which could have been far better spent tackling another question. It is important to balance the desire to finish the question with an appreciation of when 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, "Hence, or otherwise, show . . ." gives them the freedom to use any method of their choosing; often the "hence" will be the easiest, but in Question 5 this year, the "otherwise" approach was very popular.) On some questions, some candidates tried to work forwards from the given question and backwards from the answer, hoping that they would meet somewhere in the middle. While this worked on occasion, it often required fudging. 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.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1473.5
Banger Comparisons: 2
Show that
\[
\sin(x+y) -\sin(x-y) = 2 \cos x \, \sin y
\]
and deduce that
\[
\sin A - \sin B = 2
\cos \tfrac12 (A+B)
\,
\sin\tfrac12 (A-B)
\,.
\]
Show also that
\[
\cos A - \cos B =
-2
\sin \tfrac12(A+B)
\,
\sin\tfrac12(A-B)\,.
\]
The points $P$, $Q$, $R$ and $S$ have coordinates
$\left(a\cos p,b\sin p\right)$,
$\left(a\cos q,b\sin q\right)$,
$\left(a\cos r,b\sin r\right)$ and
$\left(a\cos s,b\sin s\right)$ respectively,
where $0\le p < q < r < s <2\pi$, and $a$ and $b$ are positive.
Given that neither of the lines $PQ$ and $SR$ is vertical,
show that these lines are parallel if and only if
\[
r+s-p-q = 2\pi\,.
\]\begin{align*}
&& \sin(x+y) - \sin(x-y) &= \sin x \cos y + \cos x \sin y - (\sin x \cos y - \cos x \sin y )\\
&&&= 2 \cos x \sin y \\
\\
&& A &= x+y \\
&& B &= x - y \\
\Rightarrow && x = \frac12(A+B) &\quad y = \frac12(A-B) \\
\Rightarrow && \sin A - \sin B &= 2 \cos \tfrac12(A+B) \sin \tfrac12(A-B) \\
\\
&& \cos (x+y) - \cos (x-y) &= \cos x \cos y - \sin x \sin y -(\cos x \cos y + \sin x \sin y ) \\
&&&= -2 \sin x \sin y \\
\Rightarrow && \cos A - \cos B &= - 2 \sin \tfrac12 (A+B) \sin \tfrac12 (A-B)
\end{align*}
\begin{align*}
&& \text{Gradient of }PQ &= \frac{b \sin q - b \sin p}{a \cos q - a \cos p } \\
&& \text{Gradient of }SR &= \frac{b \sin s - b \sin r}{a \cos s - a \cos r} \\
PQ \parallel SR \Rightarrow && \frac{b \sin q - b \sin p}{a \cos q - a \cos p } &= \frac{b \sin s - b \sin r}{a \cos s - a \cos r} \\
\Rightarrow && (\sin q - \sin p)(\cos s - \cos r) &= (\sin s - \sin r)(\cos q - \cos r) \\
\Rightarrow && -4 \cos \tfrac12(p+q) \sin\tfrac12(q-p) \sin \tfrac12(s+r) \sin \tfrac12(s-r) &= -4 \cos \tfrac12(s+r) \sin \tfrac12(s-r) \sin \tfrac12 (p+q) \sin\tfrac12 (q-p) \\
\Rightarrow && 0 &= \sin \tfrac12(s-r)\sin\tfrac12(p-q) \left ( \cos \tfrac12(p+q)\sin \tfrac12(s+r) - \sin \tfrac12 (p+q)\cos \tfrac12(s+r) \right) \\
&&&= \sin \tfrac12(s-r)\sin\tfrac12(p-q) \sin \left ( \frac12 (s+r -(p+q))\right)
\end{align*}
Since $s \neq r$ and $p \neq q$ (neither line vertical) we must have $\frac12 (s+r -(p+q)) = n \pi \Rightarrow s+r - p - q = 0, 2\pi, 4\pi, \cdots$ but given the range constraints it must be $2 \pi$
This was a very popular question, and the marks were generally very encouraging. A significant majority of attempts simply rattled through the first part of the question, showing confidence and competence with their trigonometric identities. A few candidates' solutions for this part lasted several pages, but the majority were very swift and efficient. For the main part of the question, most candidates started off very well by attempting to equate the lines' gradients, though a few thought that the gradient is given by the formula (x2 − x1)/(y2 − y1). A number of candidates tried to use vector methods by showing that PQ.SR/(|PQ|.|SR|) = 1, which involves significant (and messy) algebraic manipulation. (And if they used RS instead of SR, they were unlikely to try to equate to −1 instead of 1.) A few tried to show that PQ = SR, which is more restrictive than what is required. In general, the scalar (dot) product method is good for identifying perpendicular vectors, but far less useful for parallel vectors. Most candidates who followed the gradient method then reached the intermediate conclusion that tan ½(q + p) = tan ½(s + r), but did not know how to continue correctly. They either simply used the given answer or concluded that ½(q + p) + kπ = ½(s + r) where k = 0, 1 or 2. Very few said that k could be any integer or gave any justification for their restriction to the given possibilities. Finally, almost no candidates appreciated that the question's requirement to show that the lines are parallel if and only if the given condition was met meant that they had to either show that their argument was reversible (using ⇐⇒ connectives or some other indication that it was reversible), or explain why r + s − p − q = 2π implies that the lines are parallel.