Year: 2007
Paper: 2
Question Number: 4
Course: LFM Pure
Section: Trigonometry 2
Although the paper was by no means an easy one, it was generally found a more accessible paper than last year's, with most questions clearly offering candidates an attackable starting-point. The candidature represented the usual range of mathematical talents, with a pleasingly high number of truly outstanding students; many more who were able to demonstrate a thorough grasp of the material in at least three questions; and the few whose three-hour long experience was unlikely to have been a particularly pleasant one. However, even for these candidates, many were able to make some progress on at least two of the questions chosen. Really able candidates generally produced solid attempts at five or six questions, and quite a few produced outstanding efforts at up to eight questions. In general, it would be best if centres persuaded candidates not to spend valuable time needlessly in this way – it is a practice that is not to be encouraged, as it uses valuable examination time to little or no avail. Weaker brethren were often to be found scratching around at bits and pieces of several questions, with little of substance being produced on more than a couple. It is an important examination skill – now more so than ever, with most candidates now not having to employ such a skill on the modular papers which constitute the bulk of their examination experience – for candidates to spend a few minutes at some stage of the examination deciding upon their optimal selection of questions to attempt. As a rule, question 1 is intended to be accessible to all takers, with question 2 usually similarly constructed. In the event, at least one – and usually both – of these two questions were among candidates' chosen questions. These, along with questions 3 and 6, were by far the most popularly chosen questions to attempt. The majority of candidates only attempted questions in Section A (Pure Maths), and there were relatively few attempts at the Applied Maths questions in Sections B & C, with Mechanics proving the more popular of the two options. It struck me that, generally, the working produced on the scripts this year was rather better set-out, with a greater logical coherence to it, and this certainly helps the markers identify what each candidate thinks they are doing. Sadly, this general remark doesn't apply to the working produced on the Mechanics questions, such as they were. As last year, the presentation was usually appalling, with poorly labelled diagrams, often with forces missing from them altogether, and little or no attempt to state the principles that the candidates were attempting to apply.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Given that $\cos A$, $\cos B$ and $\beta$ are non-zero, show
that
the equation
\[
\alpha \sin(A-B) + \beta \cos(A+B) = \gamma \sin(A+B)
\]
reduces to the form
\[
(\tan A-m)(\tan B-n)=0\,,
\]
where $m$ and $n$ are independent of $A$ and $B$,
if and only if $\alpha^2=\beta^2+\gamma^2$.
Determine all values of $x$, in the range $0\le x <2\pi$, for which:
\begin{questionparts}
\item $2\sin(x-\frac14\pi) + \sqrt{3} \cos(x+\frac14\pi) =
\sin(x+\frac14\pi)$
\item $2\sin(x-\frac16\pi) + \sqrt{3} \cos(x+\frac16\pi) =
\sin(x+\frac16\pi)$
\item $2\sin(x+\frac13\pi) + \sqrt{3} \cos(3x) =
\sin(3x)$
\end{questionparts}\begin{align*}
&& \alpha \sin(A-B) + \beta \cos (A + B) &= \gamma \sin(A+B) \\
\Leftrightarrow && \alpha \sin A \cos B - \alpha \cos A \sin B + \beta \cos A \cos B - \beta \sin A \sin B &= \gamma \sin A \cos B + \gamma \cos A \sin B \\
\Leftrightarrow && \alpha \tan A - \alpha \tan B + \beta - \beta \tan A \tan B &= \gamma \tan A + \gamma \tan B \\
\Leftrightarrow && \beta \tan A \tan B +(\gamma-\alpha) \tan A + (\gamma +\alpha)\tan B&=\beta \\
\Leftrightarrow && \tan A \tan B +\left (\frac{\gamma-\alpha}{\beta} \right) \tan A + \left (\frac{\gamma +\alpha}{\beta} \right)\tan B&=1\\
\Leftrightarrow && \left ( \tan A + \left ( \frac{\gamma+\alpha}{\beta} \right) \right) \left ( \tan B + \left ( \frac{\gamma-\alpha}{\beta} \right)\right) - \frac{\gamma^2 - \alpha^2}{\beta^2}&=1\\
\Leftrightarrow && \left ( \tan A + \left ( \frac{\gamma+\alpha}{\beta} \right) \right) \left ( \tan B + \left ( \frac{\gamma-\alpha}{\beta} \right)\right) &= \frac{\beta^2+\gamma^2-\alpha^2}{\beta^2}\\
\end{align*}
Which has the desired form iff $\beta^2+\gamma^2 = \alpha^2$.
\begin{questionparts}
\item $\,$ \begin{align*}
&& 2\sin(x-\tfrac14\pi) + \sqrt 3 \cos(x+\tfrac14\pi) &=\sin(x+\tfrac14\pi) \\
&& 3 + 1 &= 4 \\
\Rightarrow && \left (\tan x + \frac{1+2}{\sqrt3} \right) \left ( \tan \frac{\pi}{4} + \frac{1-2}{\sqrt3}\right) &= 0\\
\Rightarrow && \tan x &= -\sqrt3 \\
\Rightarrow && x &= \tfrac23\pi, \tfrac53\pi
\end{align*}
\item $\,$ \begin{align*}
&& 2\sin(x-\frac16\pi) + \sqrt 3 \cos(x+\frac16\pi) &=\sin(x+\frac16\pi) \\
\Leftrightarrow && \left (\tan x + \frac{1+2}{\sqrt3} \right) \left ( \tan \frac{\pi}{3} + \frac{1-2}{\sqrt3}\right) &= 0\\
&& x &\in [0, 2\pi)
\end{align*}
\item $\,$ \begin{align*}
&& 2\sin(x+\frac13\pi) + \sqrt 3 \cos(3x) = \sin(3x) \\
&& A-B =x + \tfrac13\pi, A+B &= 3x \\
\Rightarrow && A = 2x + \tfrac\pi6, B &= x-\tfrac{\pi}{6} \\
\Rightarrow && \tan (2x+\tfrac\pi6)&=-\sqrt3 \\
&& 2x + \tfrac{\pi}{6} &= \tfrac23\pi, \tfrac53\pi, \tfrac83 \pi, \tfrac{11}3\pi \\
&& x &= \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4} \\
&& \tan(-x-\tfrac{\pi}{6}) &= \frac1{\sqrt{3}} \\
\Rightarrow && x-\tfrac{\pi}{6} &= \ldots, \tfrac{\pi}{6}, \tfrac{7\pi}{6}, \ldots \\
\Rightarrow && x &= \tfrac{\pi}3, \tfrac{4\pi}{3} \\
\\
\Rightarrow && x &= \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4} , \tfrac{\pi}3, \tfrac{4\pi}{3}
\end{align*}
\end{questionparts}
This question was a popular one for partial attempts; with most candidates giving up towards the end of the introductory part and going elsewhere. It was slightly surprising to see candidates being put off in this way, since the given result made it perfectly possible to move successfully into the three following cases. For those who did press on, many lost a mark for not verifying (somewhere) that the chosen values of α, β and γ actually satisfied the required condition. Then, in (ii), one of the two brackets was identically zero, the significance of which was largely overlooked, with many candidates offering again the same two solutions as had been found in part (i). In (iii), it was important to note first A and B in terms of x, although some candidates adopted a valid alternative approach by first collecting up the two 3x terms.