| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | January |
| Marks | 12 |
| 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 standard C2 trigonometric equations question with routine techniques: part (a) uses periodicity of tan, part (b) involves algebraic manipulation using sin²θ + cos²θ = 1 and solving a quadratic, then applying to a double angle. All steps are textbook procedures 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=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x + 52° = (22°),\ 180° + 22°;\ 360° + 22°\) \((x = 180+22-52;\ x = 360+22-52)\) | M1;M1 | \(x + 52 = 180 +\) AWRT 22, \(360 +\) AWRT 22 OE. (max of M1 if extras in range). LHS could be any letter but not \(x\) unless final answer shows recovery. Ms can be PI |
| \(x = 150°,\ 330°\) | A1 | Both CAO with no extras in \(0° \leq x \leq 360°\). Ignore anything outside \(0° \leq x \leq 360°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(3\tan\theta = \frac{8}{\sin\theta} \Rightarrow 3\frac{\sin\theta}{\cos\theta} = \frac{8}{\sin\theta}\) | M1 | \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) used/seen |
| \(\frac{3(1-\cos^2\theta)}{\cos\theta} = 8\) | M1 | \(\sin^2\theta = 1 - \cos^2\theta\) used |
| \(\Rightarrow 3 - 3\cos^2\theta = 8\cos\theta\) \(\Rightarrow 3\cos^2\theta + 8\cos\theta - 3 = 0\) | A1 | CSO AG Completion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((3\cos\theta - 1)(\cos\theta + 3) = 0\) | M1 | Any valid method to solve the quadratic |
| \(\cos\theta = \frac{1}{3}\) | A1 | CSO Must only be the one value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\cos 2x = \frac{1}{3}\) | M1 | Using (ii) OE to get or use \(\cos 2x = k\) where \(-1 \leq k \leq 1\) |
| \((2x =)\ 70.528\ldots\) | B1 | Award for \(\cos^{-1}(1/3)\) = value from 70 to 71 inclusive, even if \(\theta\) used. PI |
| \(2x = 360° - 70.528\ldots\ (= 289.47\ldots)\) | m1 | \(2x = 360 - \cos^{-1}(c\text{'s } k)\) OE. No extras inside the range |
| \(x = 35°,\ 145°\) (to the nearest degree) | A1 | Both, condoning greater accuracy, with no extras in \(0° \leq x \leq 180°\). Ignore anything outside \(0° \leq x \leq 180°\). SC for (b)(iii) only when \(c\)'s answer for (b)(ii) is \(\cos\theta = -\frac{1}{3}\): max mark M1B1 (val 70–71 or val 109–110 inclusive) m1A0 |
## Question 8:
### Part 8(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x + 52° = (22°),\ 180° + 22°;\ 360° + 22°$ $(x = 180+22-52;\ x = 360+22-52)$ | M1;M1 | $x + 52 = 180 +$ AWRT 22, $360 +$ AWRT 22 OE. (max of M1 if extras in range). LHS could be any letter but not $x$ unless final answer shows recovery. Ms can be PI |
| $x = 150°,\ 330°$ | A1 | Both CAO with no extras in $0° \leq x \leq 360°$. Ignore anything outside $0° \leq x \leq 360°$ |
### Part 8(b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $3\tan\theta = \frac{8}{\sin\theta} \Rightarrow 3\frac{\sin\theta}{\cos\theta} = \frac{8}{\sin\theta}$ | M1 | $\tan\theta = \frac{\sin\theta}{\cos\theta}$ used/seen |
| $\frac{3(1-\cos^2\theta)}{\cos\theta} = 8$ | M1 | $\sin^2\theta = 1 - \cos^2\theta$ **used** |
| $\Rightarrow 3 - 3\cos^2\theta = 8\cos\theta$ $\Rightarrow 3\cos^2\theta + 8\cos\theta - 3 = 0$ | A1 | CSO AG Completion |
### Part 8(b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(3\cos\theta - 1)(\cos\theta + 3) = 0$ | M1 | Any valid method to solve the quadratic |
| $\cos\theta = \frac{1}{3}$ | A1 | CSO Must only be the one value |
### Part 8(b)(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\cos 2x = \frac{1}{3}$ | M1 | Using (ii) OE to get or use $\cos 2x = k$ where $-1 \leq k \leq 1$ |
| $(2x =)\ 70.528\ldots$ | B1 | Award for $\cos^{-1}(1/3)$ = value from 70 to 71 inclusive, even if $\theta$ used. PI |
| $2x = 360° - 70.528\ldots\ (= 289.47\ldots)$ | m1 | $2x = 360 - \cos^{-1}(c\text{'s } k)$ OE. No extras inside the range |
| $x = 35°,\ 145°$ (to the nearest degree) | A1 | Both, condoning greater accuracy, with no extras in $0° \leq x \leq 180°$. Ignore anything outside $0° \leq x \leq 180°$. **SC for (b)(iii) only when $c$'s answer for (b)(ii) is $\cos\theta = -\frac{1}{3}$:** max mark M1B1 (val 70–71 or val 109–110 inclusive) m1A0 |8
\begin{enumerate}[label=(\alph*)]
\item Solve the equation $\tan \left( x + 52 ^ { \circ } \right) = \tan 22 ^ { \circ }$, giving the values of $x$ in the interval $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$.
\item \begin{enumerate}[label=(\roman*)]
\item Show that the equation
$$3 \tan \theta = \frac { 8 } { \sin \theta }$$
can be written as
$$3 \cos ^ { 2 } \theta + 8 \cos \theta - 3 = 0$$
\item Find the value of $\cos \theta$ that satisfies the equation
$$3 \cos ^ { 2 } \theta + 8 \cos \theta - 3 = 0$$
\item Hence solve the equation
$$3 \tan 2 x = \frac { 8 } { \sin 2 x }$$
giving all values of $x$ to the nearest degree in the interval $0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2010 Q8 [12]}}