| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Show then solve substituted equation |
| Difficulty | Standard +0.3 This is a structured two-part question where part (a) guides students through algebraic manipulation using the Pythagorean identity (sin²θ + cos²θ = 1) to reach a given answer, and part (b) applies this result to a compound angle case requiring straightforward substitution and angle division. The algebraic manipulation is routine for C2 level, and the compound angle solving is standard practice, making this slightly easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{3+\sin^2\theta}{\cos\theta - 2} = 3\cos\theta \Rightarrow \frac{3+(1-\cos^2\theta)}{\cos\theta-2} = 3\cos\theta\) | M1 | \(\cos^2\theta + \sin^2\theta = 1\) stated or used |
| \(\Rightarrow \frac{4-\cos^2\theta}{\cos\theta-2} = 3\cos\theta \Rightarrow \frac{(2-\cos\theta)(2+\cos\theta)}{\cos\theta-2} = 3\cos\theta\) | m1 | Difference of two squares, or division (PI by next line) |
| \(\Rightarrow -1(2+\cos\theta) = 3\cos\theta\) | A1 | |
| \(\Rightarrow -2 = 4\cos\theta \Rightarrow \cos\theta = -\frac{1}{2}\) | A1 | CSO AG; total 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3+1-\cos^2\theta = 3\cos^2\theta - 6\cos\theta\) | (M1) | \(\cos^2\theta + \sin^2\theta = 1\) |
| \((4\cos\theta+2)(\cos\theta-2)=0\) | (m1) | Factorising or formula |
| \(\cos\theta - 2 \neq 0\) | (A1) | Indicates rejection of \(\cos\theta = 2\) |
| \(\Rightarrow 4\cos\theta = -2 \Rightarrow \cos\theta = -\frac{1}{2}\) | (A1) | AG Be convinced |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\theta = 3x \Rightarrow \cos 3x = -\frac{1}{2}\) | M1 | Uses part (a) to reach either \(\cos 3x = -0.5\) or \(\cos 3x = 0.5\) |
| \(\cos^{-1}\!\left(-\frac{1}{2}\right) = 120°\) | m1 | Or \(\cos^{-1}(0.5) = 60°\). Condone radians here |
| \(3x = 120°, 240°, 480°, \ldots\) | ||
| \(x = 40°, 80°, 160°\) | A2,1,0 | A1 for at least two correct. If \(>3\) solutions in \(0° < x < 180°\), deduct 1 mark from any A marks for each extra solution. Deduct 1 mark from any A marks if answers in radians. Ignore extra values outside the given interval; total 4 marks |
## Question 9:
**Part (a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{3+\sin^2\theta}{\cos\theta - 2} = 3\cos\theta \Rightarrow \frac{3+(1-\cos^2\theta)}{\cos\theta-2} = 3\cos\theta$ | M1 | $\cos^2\theta + \sin^2\theta = 1$ stated or used |
| $\Rightarrow \frac{4-\cos^2\theta}{\cos\theta-2} = 3\cos\theta \Rightarrow \frac{(2-\cos\theta)(2+\cos\theta)}{\cos\theta-2} = 3\cos\theta$ | m1 | Difference of two squares, or division (PI by next line) |
| $\Rightarrow -1(2+\cos\theta) = 3\cos\theta$ | A1 | |
| $\Rightarrow -2 = 4\cos\theta \Rightarrow \cos\theta = -\frac{1}{2}$ | A1 | CSO AG; total 4 marks |
**Alternative for (a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3+1-\cos^2\theta = 3\cos^2\theta - 6\cos\theta$ | (M1) | $\cos^2\theta + \sin^2\theta = 1$ |
| $(4\cos\theta+2)(\cos\theta-2)=0$ | (m1) | Factorising or formula |
| $\cos\theta - 2 \neq 0$ | (A1) | Indicates rejection of $\cos\theta = 2$ |
| $\Rightarrow 4\cos\theta = -2 \Rightarrow \cos\theta = -\frac{1}{2}$ | (A1) | AG Be convinced |
**Part (b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\theta = 3x \Rightarrow \cos 3x = -\frac{1}{2}$ | M1 | Uses part (a) to reach either $\cos 3x = -0.5$ or $\cos 3x = 0.5$ |
| $\cos^{-1}\!\left(-\frac{1}{2}\right) = 120°$ | m1 | Or $\cos^{-1}(0.5) = 60°$. Condone radians here |
| $3x = 120°, 240°, 480°, \ldots$ | | |
| $x = 40°, 80°, 160°$ | A2,1,0 | A1 for at least two correct. If $>3$ solutions in $0° < x < 180°$, deduct 1 mark from any A marks for each extra solution. Deduct 1 mark from any A marks if answers in radians. Ignore extra values outside the given interval; total 4 marks |9
\begin{enumerate}[label=(\alph*)]
\item Given that
$$\frac { 3 + \sin ^ { 2 } \theta } { \cos \theta - 2 } = 3 \cos \theta$$
show that
$$\cos \theta = - \frac { 1 } { 2 }$$
\item Hence solve the equation
$$\frac { 3 + \sin ^ { 2 } 3 x } { \cos 3 x - 2 } = 3 \cos 3 x$$
giving all solutions in degrees in the interval $0 ^ { \circ } < x < 180 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2008 Q9 [8]}}