| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Solve tan·sin or tan·trig product |
| Difficulty | Moderate -0.3 This is a structured, multi-part C2 trigonometric equation question that guides students through each step: converting using tan=sin/cos (1 mark), applying the Pythagorean identity (standard), solving a quadratic, rejecting an invalid solution based on cosine range, and finding angles. The scaffolding makes it easier than average, though part (c) requires understanding the transformation to 4x. Overall slightly below average difficulty for A-level. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| 8(a)(i) | \(\frac{4\sin\theta}{\cos\theta}\sin\theta = 15\) ⟹ \(4\sin^2\theta = 15\cos\theta\) | B1 |
| 8(a)(ii) | \(\sin^2\theta + \cos^2\theta = 1\); \(4(1 - \cos^2\theta) = 15\cos\theta\); \(4\cos^2\theta + 15\cos\theta - 4 = 0\) | M1, A1 |
| 8(b)(i) | \((4c - 1)(c + 4) = 0\); \(c = -4, c = \frac{1}{4}\) | M1, A1 |
| 8(b)(ii) | Since \(-1 ≤ \cos\theta ≤ \{1\}\), the only possible value for \(\cos\theta\) is \(\frac{1}{4}\) | E1 ft |
| 8(b)(iii) | \(\theta = 75.5°\); \(\theta = 284.5°\) | B1, B1 ft |
| 8(c) | .......... ⟹ \(\cos 4x = \frac{1}{4}\) | M1 |
| \(x = 19°, 71°\) | A1 ft | Ft on (iii)4......(only ft if 2 answers in given range). |
8(a)(i) | $\frac{4\sin\theta}{\cos\theta}\sin\theta = 15$ ⟹ $4\sin^2\theta = 15\cos\theta$ | B1 | AG Be convinced |
8(a)(ii) | $\sin^2\theta + \cos^2\theta = 1$; $4(1 - \cos^2\theta) = 15\cos\theta$; $4\cos^2\theta + 15\cos\theta - 4 = 0$ | M1, A1 | OE seen; AG Be convinced |
8(b)(i) | $(4c - 1)(c + 4) = 0$; $c = -4, c = \frac{1}{4}$ | M1, A1 | Factorisation or formula or completion of square; Both values |
8(b)(ii) | Since $-1 ≤ \cos\theta ≤ \{1\}$, the only possible value for $\cos\theta$ is $\frac{1}{4}$ | E1 ft | AG convincingly explained (Condone strict inequalities); Ft provided candidates answers for c are $\frac{1}{4}$ and a value k such that k > 1 or k < −1 |
8(b)(iii) | $\theta = 75.5°$; $\theta = 284.5°$ | B1, B1 ft | Ft on [360 − c's 75.5°] as only other solution in the given interval |
8(c) | .......... ⟹ $\cos 4x = \frac{1}{4}$ | M1 | Links with previous parts. PI |
| $x = 19°, 71°$ | A1 ft | Ft on (iii)4......(only ft if 2 answers in given range). |8
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the equation
$$4 \tan \theta \sin \theta = 15$$
can be written as
$$4 \sin ^ { 2 } \theta = 15 \cos \theta$$
(1 mark)
\item Use an appropriate identity to show that the equation
$$4 \sin ^ { 2 } \theta = 15 \cos \theta$$
can be written as
$$4 \cos ^ { 2 } \theta + 15 \cos \theta - 4 = 0$$
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Solve the equation $4 c ^ { 2 } + 15 c - 4 = 0$.
\item Hence explain why the only value of $\cos \theta$ which satisfies the equation
$$4 \cos ^ { 2 } \theta + 15 \cos \theta - 4 = 0$$
is $\cos \theta = \frac { 1 } { 4 }$.
\item Hence solve the equation $4 \tan \theta \sin \theta = 15$ giving all solutions to the nearest $0.1 ^ { \circ }$ in the interval $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.
\end{enumerate}\item Write down all the values of $x$ in the interval $0 ^ { \circ } \leqslant x \leqslant 90 ^ { \circ }$ for which
$$4 \tan 4 x \sin 4 x = 15$$
giving your answers to the nearest degree.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2005 Q8 [10]}}