Year: 2007
Paper: 1
Question Number: 2
Course: LFM Pure
Section: Trigonometry 2
There were significantly more candidates attempting this paper this year (an increase of nearly 50%), but many found it to be very difficult and only achieved low scores. In particular, the level of algebraic skill required by the questions was often lacking. The examiners' express their concern that this was the case despite a conscious effort to make the paper more accessible than last year's. At this level, the fluent, confident and correct handling of mathematical symbols (and numbers) is necessary and is expected; many good starts to questions soon became unstuck after a simple slip. Graph sketching was usually poor: if future candidates wanted to improve one particular skill, they would be well advised to develop this. There were of course some excellent scripts, full of logical clarity and perceptive insight. It was pleasing to note that the applied questions were more popular this year, and many candidates scored well on at least one of these. It was however surprising how rarely answers to questions such as 5, 9, 10, 11 and 12 began with a diagram. However, the examiners were left with the overall feeling that some candidates had not prepared themselves well for the examination. The use of past papers to ensure adequate preparation is strongly recommended. A student's first exposure to STEP questions can be a daunting, demanding experience; it is a shame if that takes place during a public examination on which so much rides. Further, and fuller, discussion of the solutions to these questions can be found in the Hints and Answers document.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1515.7
Banger Comparisons: 3
\begin{questionparts}
\item Given that $A = \arctan \frac12$ and that $B = \arctan\frac13\,$ (where $A$ and $B$ are acute) show, by considering $\tan \left( A + B \right)$, that $A + B = {\frac{1}{4}\pi }$.
The non-zero integers $p$ and $q$ satisfy
\[
\displaystyle \arctan {\frac1 p} + \arctan {\frac1 q}
= {\frac\pi 4}\,.
\]
Show that $ \left ( p-1 \right) \left(q-1 \right) = 2$ and hence determine $p$ and $q$.
\item Let $r$, $s$ and $t$ be positive integers such that the highest common factor of $s$ and $t$ is $1$. Show that, if
\[
\arctan {\frac1 r} + \arctan \frac s {s+t} = {\frac\pi 4}\,,
\]
then there are only two possible values for $t$, and give $r$ in terms of $s$ in each case.
\end{questionparts}\begin{questionparts}
\item \begin{align*}
&& \tan (A+B) &= \frac{\tan A + \tan B}{1-\tan A \tan B}\\
&&&= \frac{\tan \arctan \frac12 + \tan \arctan \frac13}{1-\tan \arctan \frac12 \tan \arctan \frac13}\\
&&&= \frac{\frac12+\frac13}{1-\frac16} \\
&&&= \frac{3+2}{5} \\
&&&= 1 \\
\Rightarrow && A+B &= \frac{\pi}{4} + n \pi
\end{align*}
but since $A,B$ are acute $0 < A+B < \pi$, so $A+B = \frac{\pi}{4}$
\begin{align*}
&& 1 &= \tan \frac{\pi}{4} \\
&&&= \tan \left ( \arctan {\frac1 p} + \arctan {\frac1 q}\right) \\
&&&= \frac{\frac1p + \frac1q}{1-\frac1{pq}} \\
&&&= \frac{q+p}{pq-1} \\
\Rightarrow && pq-1 &= q+p \\
\Rightarrow && 0 &= pq-q-p-q \\
&&&= (p-1)(q-1)-2 \\
\Rightarrow && 2 &= (p-1)(q-1)
\end{align*}
But $p$,$q$ are integers, so $p-1 \in \{-2,-1,1,1\} \Rightarrow p \in \{-1,0,2,3\}$ but we cannot have $p= 0$, so we must have
$(p,q) = (2,3), (3,2)$
\item \begin{align*}
&& 1 &= \tan \frac{\pi}{4} \\
&&&= \tan \left ( \arctan {\frac1 r} + \arctan \frac s {s+t} \right) \\
&&&= \frac{\frac1r + \frac{s}{s+t}}{1-\frac{s}{r(s+t)}} \\
&&&= \frac{s+t+sr}{r(s+t)-s} \\
\Rightarrow && rs+rt-s &= s+t + sr \\
\Rightarrow && 0 &= rt-2s-t \\
&&2s&= t(r-1)
\end{align*}
Since $(s,t) =1$, we must have $t \mid 2$, so $ t = 1,2$ and $r = 2s+1$ or $r=s+1$ respectively.
\end{questionparts}
This was a popular question, and was usually well done. The argument at the end was often incomplete, though: many candidates simply stated that t = 1 or t = 2 without explaining why no other values were possible. To do so, use had to be made of the fact that s and t have no common factor other than 1.