| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.3 This is a standard C3 trigonometric identity question with routine algebraic manipulation. Part (a) requires combining fractions over a common denominator and simplifying—a textbook exercise. Part (b) uses the result to solve a quadratic in sec x, requiring standard techniques. Part (c) is a simple substitution. While multi-step, it follows predictable patterns without requiring novel insight, making it slightly easier than average. |
| Spec | 1.05j 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 |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Common denominator: \(\frac{(1-\sin x)^2 + \cos^2 x}{\cos x(1-\sin x)}\) | M1 | Correct combination |
| Numerator: \(1 - 2\sin x + \sin^2 x + \cos^2 x = 2 - 2\sin x\) | M1 A1 | Using \(\sin^2 x + \cos^2 x = 1\) |
| \(= \frac{2(1-\sin x)}{\cos x(1-\sin x)} = \frac{2}{\cos x} = 2\sec x\) | A1 | Completion of proof |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(2\sec x = \sec^2 x - 1 - 2 = \sec^2 x - 3\) | M1 | Using \(\tan^2 x = \sec^2 x - 1\) |
| \(\sec^2 x - 2\sec x - 3 = 0\) | M1 | Forming quadratic in \(\sec x\) |
| \((\sec x - 3)(\sec x + 1) = 0\) | M1 | Factorising |
| \(\sec x = 3 \Rightarrow \cos x = \frac{1}{3}\); \(\sec x = -1 \Rightarrow \cos x = -1\) | A1 | Both values |
| \(x = 180°\) (from \(\cos x = -1\)) | A1 | |
| \(x \approx 70.5°\), \(289.5°\) (from \(\cos x = \frac{1}{3}\)) | A1 | Both values, to nearest degree |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Replace \(x\) with \(2\theta - 30°\), solutions from (b) give \(2\theta - 30° = 70.5°, 180°, 289.5°\) | M1 | Correct substitution |
| \(\theta = 50°, 105°\) (values in range \(0° \leq \theta \leq 180°\)) | A1 | Both correct values |
# Question 8(a):
Show $\frac{1-\sin x}{\cos x} + \frac{\cos x}{1-\sin x} = 2\sec x$
| Working/Answer | Mark | Guidance |
|---|---|---|
| Common denominator: $\frac{(1-\sin x)^2 + \cos^2 x}{\cos x(1-\sin x)}$ | M1 | Correct combination |
| Numerator: $1 - 2\sin x + \sin^2 x + \cos^2 x = 2 - 2\sin x$ | M1 A1 | Using $\sin^2 x + \cos^2 x = 1$ |
| $= \frac{2(1-\sin x)}{\cos x(1-\sin x)} = \frac{2}{\cos x} = 2\sec x$ | A1 | Completion of proof |
---
# Question 8(b):
$2\sec x = \tan^2 x - 2$
| Working/Answer | Mark | Guidance |
|---|---|---|
| $2\sec x = \sec^2 x - 1 - 2 = \sec^2 x - 3$ | M1 | Using $\tan^2 x = \sec^2 x - 1$ |
| $\sec^2 x - 2\sec x - 3 = 0$ | M1 | Forming quadratic in $\sec x$ |
| $(\sec x - 3)(\sec x + 1) = 0$ | M1 | Factorising |
| $\sec x = 3 \Rightarrow \cos x = \frac{1}{3}$; $\sec x = -1 \Rightarrow \cos x = -1$ | A1 | Both values |
| $x = 180°$ (from $\cos x = -1$) | A1 | |
| $x \approx 70.5°$, $289.5°$ (from $\cos x = \frac{1}{3}$) | A1 | Both values, to nearest degree |
---
# Question 8(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Replace $x$ with $2\theta - 30°$, solutions from (b) give $2\theta - 30° = 70.5°, 180°, 289.5°$ | M1 | Correct substitution |
| $\theta = 50°, 105°$ (values in range $0° \leq \theta \leq 180°$) | A1 | Both correct values |
The images you've shared show only blank answer space pages (pages 17-20) from what appears to be an AQA Mathematics exam paper (P/Jun14/MPC3). These pages contain no mark scheme content — they are simply lined answer spaces for Question 8, followed by an "END OF QUESTIONS" page and a blank page.
To get the mark scheme content, you would need to share the actual **mark scheme document** for this paper, not the question paper answer pages.
If you share the mark scheme pages, I'd be happy to extract and format the content as requested.8
\begin{enumerate}[label=(\alph*)]
\item Show that the expression $\frac { 1 - \sin x } { \cos x } + \frac { \cos x } { 1 - \sin x }$ can be written as $2 \sec x$.\\[0pt]
[4 marks]
\item Hence solve the equation
$$\frac { 1 - \sin x } { \cos x } + \frac { \cos x } { 1 - \sin x } = \tan ^ { 2 } x - 2$$
giving the values of $x$ to the nearest degree in the interval $0 ^ { \circ } \leqslant x < 360 ^ { \circ }$.\\[0pt]
[6 marks]
\item Hence solve the equation
$$\frac { 1 - \sin \left( 2 \theta - 30 ^ { \circ } \right) } { \cos \left( 2 \theta - 30 ^ { \circ } \right) } + \frac { \cos \left( 2 \theta - 30 ^ { \circ } \right) } { 1 - \sin \left( 2 \theta - 30 ^ { \circ } \right) } = \tan ^ { 2 } \left( 2 \theta - 30 ^ { \circ } \right) - 2$$
giving the values of $\theta$ to the nearest degree in the interval $0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }$.\\[0pt]
[2 marks]
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{57412ec0-ad97-4418-8ba8-93f1f7d8aac1-16_1517_1709_1190_153}
\end{center}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{57412ec0-ad97-4418-8ba8-93f1f7d8aac1-20_2489_1730_221_139}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2014 Q8 [12]}}