| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Convert equation to quadratic form |
| Difficulty | Standard +0.3 Part (a) is a straightforward algebraic simplification using the difference of squares and the Pythagorean identity sec²x - 1 = tan²x. Part (b) requires converting to a quadratic in sin x and solving, which is standard C3 technique. Part (c) is a direct application of part (b) with a compound angle substitution. This is a typical multi-part C3 question that tests standard identities and equation-solving techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \((\sec x+1)(\sec x-1) = \sec^2 x - 1\) | M1 | Expand denominator |
| Use \(\sec^2 x - 1 = \tan^2 x\) | M1 | Apply identity |
| \(\frac{\sec^2 x}{\tan^2 x} = \frac{1}{\sin^2 x} = \cosec^2 x\) | A1 | Complete proof using \(\frac{\sec^2 x}{\tan^2 x} = \frac{1/\cos^2 x}{\sin^2 x/\cos^2 x}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\cosec^2 x = \cosec x + 3\) | M1 | Use part (a) result |
| \(\cosec^2 x - \cosec x - 3 = 0\) | M1 | Rearrange to quadratic in \(\cosec x\) |
| \(\cosec x = \frac{1 \pm \sqrt{13}}{2}\) | M1 A1 | Solve quadratic |
| \(\sin x = \frac{2}{1+\sqrt{13}}\) giving \(x \approx 31°, 149°\) | A1 | |
| \(\sin x = \frac{2}{1-\sqrt{13}}\) giving \(x \approx -22°, -158°\) | A1 | Both pairs needed |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Replace \(x\) with \(2\theta - 60°\), use part (b) values | M1 | e.g. \(2\theta - 60° = 31°\) or \(-22°\) |
| \(\theta = 19°\) and \(\theta = 45°\) (values in range \(0° < \theta < 90°\)) | A1 | Both needed, ignore out-of-range |
## Question 6:
**Part (a):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $(\sec x+1)(\sec x-1) = \sec^2 x - 1$ | M1 | Expand denominator |
| Use $\sec^2 x - 1 = \tan^2 x$ | M1 | Apply identity |
| $\frac{\sec^2 x}{\tan^2 x} = \frac{1}{\sin^2 x} = \cosec^2 x$ | A1 | Complete proof using $\frac{\sec^2 x}{\tan^2 x} = \frac{1/\cos^2 x}{\sin^2 x/\cos^2 x}$ |
**Part (b):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\cosec^2 x = \cosec x + 3$ | M1 | Use part (a) result |
| $\cosec^2 x - \cosec x - 3 = 0$ | M1 | Rearrange to quadratic in $\cosec x$ |
| $\cosec x = \frac{1 \pm \sqrt{13}}{2}$ | M1 A1 | Solve quadratic |
| $\sin x = \frac{2}{1+\sqrt{13}}$ giving $x \approx 31°, 149°$ | A1 | |
| $\sin x = \frac{2}{1-\sqrt{13}}$ giving $x \approx -22°, -158°$ | A1 | Both pairs needed |
**Part (c):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Replace $x$ with $2\theta - 60°$, use part (b) values | M1 | e.g. $2\theta - 60° = 31°$ or $-22°$ |
| $\theta = 19°$ and $\theta = 45°$ (values in range $0° < \theta < 90°$) | A1 | Both needed, ignore out-of-range |
---6
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { \sec ^ { 2 } x } { ( \sec x + 1 ) ( \sec x - 1 ) }$$
can be written as $\operatorname { cosec } ^ { 2 } x$.
\item Hence solve the equation
$$\frac { \sec ^ { 2 } x } { ( \sec x + 1 ) ( \sec x - 1 ) } = \operatorname { cosec } x + 3$$
giving the values of $x$ to the nearest degree in the interval $- 180 ^ { \circ } < x < 180 ^ { \circ }$.
\item Hence solve the equation
$$\frac { \sec ^ { 2 } \left( 2 \theta - 60 ^ { \circ } \right) } { \left( \sec \left( 2 \theta - 60 ^ { \circ } \right) + 1 \right) \left( \sec \left( 2 \theta - 60 ^ { \circ } \right) - 1 \right) } = \operatorname { cosec } \left( 2 \theta - 60 ^ { \circ } \right) + 3$$
giving the values of $\theta$ to the nearest degree in the interval $0 ^ { \circ } < \theta < 90 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2013 Q6 [11]}}