| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Solve tan·sin or tan·trig product |
| Difficulty | Standard +0.3 This is a standard C2 trigonometric equation question with routine steps: sketching tan x, solving a basic equation, algebraic manipulation using tan θ = sin θ/cos θ to form a quadratic in cos θ, then applying the result to a compound angle. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Graph shown | B1 | Ignore any part of the graph drawn outside interval 0°≤x≤360° in (a) |
| B1 | A 3 branch curve between 0 and 360 meeting the x-axis at or very close to 0, 180, 360 only | |
| B1 | A 3 branch curve between 0 and 360 with correct shape tending to infinity at, at least 3, of the 4 relevant ends | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(135°;\ 315°\) | B2, 1, 0 | B2 for both 135 and 315 and no 'extras' in interval \(0° \leq x \leq 360°\) (If not B2 then award B1 for either 135 or 315 with or without extras) |
| Answer | Marks | Guidance |
|---|---|---|
| \(6\tan\theta\sin\theta = 5 \Rightarrow 6\frac{\sin\theta}{\cos\theta}\sin\theta = 5\) | M1 | \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) used |
| \(6\frac{\sin^2\theta}{\cos\theta} = 5 \Rightarrow 6\frac{1 - \cos^2\theta}{\cos\theta} = 5\) | m1 | \(\sin^2\theta\) replaced by \(1 - \cos^2\theta\) throughout |
| \(6 - 6\cos^2\theta = 5\cos\theta \Rightarrow 6\cos^2\theta + 5\cos\theta - 6 = 0\) | A1 | Completion AG Be convinced |
| Answer | Marks | Guidance |
|---|---|---|
| \(6\tan 3x\sin 3x = 5 \Rightarrow 6\cos^2 3x + 5\cos 3x - 6 = 0\) | M1 | Using (b)(i) with \(\theta = 3x\) PI by attempting to solve eg for theta then dividing soln(s) by 3 |
| \(\{3\cos 3x - 2\}(2\cos 3x + 3) (= 0)\) | m1 | Correct factorisation or correct subst into the quadratic formula PI by two "correct" roots |
| \(\{\cos 3x = \frac{2}{3}, -\frac{3}{2}\}\) | ||
| \(\cos 3x = \frac{2}{3} = \cos 48.1(89...) [= \cos\alpha]\) | m1 | Dep on M1 only. \(3x = \alpha, 360°-\alpha, 360°+\alpha\) for c's \(\alpha\). from an eqn \(\cos 3x = k\) where \(-1 < k < 1\) OE PI and no solns from \(k\) outside \(-1≤k≤1\) |
| \(3x = \alpha, 360°-\alpha, 360°+\alpha\) | ||
| \(x = 16°,\) | B1 | AWRT 16, 104, 136. Deduct one mark (from any award of these 3 B marks) if more than three solns given inside the interval 0°≤x≤180°. Ignore any solutions outside the interval 0°≤x≤180°. NMS Max. is B3/6 |
| \(104°,\) | B1 | |
| \(136°\) | B1 | |
| Total | 14 | |
| TOTAL | 75 |
**9(a)(i)**
Graph shown | B1 | Ignore any part of the graph drawn outside interval 0°≤x≤360° in (a)
| B1 | A 3 branch curve between 0 and 360 meeting the x-axis at or very close to 0, 180, 360 only
| B1 | A 3 branch curve between 0 and 360 with correct shape tending to infinity at, at least 3, of the 4 relevant ends | 3
**9(a)(ii)**
$135°;\ 315°$ | B2, 1, 0 | B2 for both 135 and 315 and no 'extras' in interval $0° \leq x \leq 360°$ (If not B2 then award B1 for either 135 or 315 with or without extras) | 2
**9(b)(i)**
$6\tan\theta\sin\theta = 5 \Rightarrow 6\frac{\sin\theta}{\cos\theta}\sin\theta = 5$ | M1 | $\tan\theta = \frac{\sin\theta}{\cos\theta}$ used
$6\frac{\sin^2\theta}{\cos\theta} = 5 \Rightarrow 6\frac{1 - \cos^2\theta}{\cos\theta} = 5$ | m1 | $\sin^2\theta$ replaced by $1 - \cos^2\theta$ throughout
$6 - 6\cos^2\theta = 5\cos\theta \Rightarrow 6\cos^2\theta + 5\cos\theta - 6 = 0$ | A1 | Completion AG Be convinced | 3
**9(b)(ii)**
$6\tan 3x\sin 3x = 5 \Rightarrow 6\cos^2 3x + 5\cos 3x - 6 = 0$ | M1 | Using (b)(i) with $\theta = 3x$ PI by attempting to solve eg for theta then dividing soln(s) by 3
$\{3\cos 3x - 2\}(2\cos 3x + 3) (= 0)$ | m1 | Correct factorisation or correct subst into the quadratic formula PI by two "correct" roots
$\{\cos 3x = \frac{2}{3}, -\frac{3}{2}\}$ | |
$\cos 3x = \frac{2}{3} = \cos 48.1(89...) [= \cos\alpha]$ | m1 | Dep on M1 only. $3x = \alpha, 360°-\alpha, 360°+\alpha$ for c's $\alpha$. from an eqn $\cos 3x = k$ where $-1 < k < 1$ OE PI and no solns from $k$ outside $-1≤k≤1$
$3x = \alpha, 360°-\alpha, 360°+\alpha$ | |
$x = 16°,$ | B1 | AWRT 16, 104, 136. Deduct one mark (from any award of these 3 B marks) if more than three solns given inside the interval 0°≤x≤180°. Ignore any solutions outside the interval 0°≤x≤180°. NMS Max. is B3/6
$104°,$ | B1 |
$136°$ | B1 | | 6
Total | | 14
**TOTAL** | | 759
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item On the axes given below, sketch the graph of $y = \tan x$ for $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$.
\item Solve the equation $\tan x = - 1$, giving all values of $x$ in the interval $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Given that $6 \tan \theta \sin \theta = 5$, show that $6 \cos ^ { 2 } \theta + 5 \cos \theta - 6 = 0$.
\item Hence solve the equation $6 \tan 3 x \sin 3 x = 5$, giving all values of $x$ to the nearest degree in the interval $0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{f4f090a1-7e36-4993-a49e-b6e7e8589057-5_720_1367_806_390}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2013 Q9 [14]}}