| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2005 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Convert equation to quadratic form |
| Difficulty | Moderate -0.3 This is a structured, multi-part question that guides students through converting a trigonometric equation using the standard identity tan²x = sec²x - 1, then solving a quadratic in sec x. While it requires knowledge of reciprocal trig identities and solving quadratics, the scaffolding makes it slightly easier than average, requiring mainly careful algebraic manipulation rather than problem-solving insight. |
| 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 | Marks | Guidance |
| \(\tan^2 x = \sec x + 11\); \(\sec^2 x - 1 = \sec x + 11\) | M1 | Or attempt to form quadratic in \(\cos^2\); \(\tan^2 x = \sec^2 x - 1\) |
| \(\sec^2 x - \sec x - 12 = 0\) | A1 | AG; Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \((\sec x - 4)(\sec x + 3) = 0\) | M1 | Attempt at solving quadratic |
| \(\sec x = 4,\,-3\) | A1F | |
| \(\therefore \cos x = \frac{1}{4},\,-\frac{1}{3}\) | A1 | AG; A0 if no use of \(\cos x = \frac{1}{\sec x}\); Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(x = 76°,\, 284°\) | B1 | 2 correct |
| \(x = 109°,\, 251°\) (or better) | B1, B1 | Total: 3 marks. \(-1\) each extra in range. If radians: \(x = 1.32, 4.97, 1.91, 4.37\); B1 any 2 correct, B1 other 2 correct |
## Question 4:
### Part (a)
| Working | Marks | Guidance |
|---------|-------|---------|
| $\tan^2 x = \sec x + 11$; $\sec^2 x - 1 = \sec x + 11$ | M1 | Or attempt to form quadratic in $\cos^2$; $\tan^2 x = \sec^2 x - 1$ |
| $\sec^2 x - \sec x - 12 = 0$ | A1 | AG; Total: 2 marks |
### Part (b)
| Working | Marks | Guidance |
|---------|-------|---------|
| $(\sec x - 4)(\sec x + 3) = 0$ | M1 | Attempt at solving quadratic |
| $\sec x = 4,\,-3$ | A1F | |
| $\therefore \cos x = \frac{1}{4},\,-\frac{1}{3}$ | A1 | AG; A0 if no use of $\cos x = \frac{1}{\sec x}$; Total: 3 marks |
### Part (c)
| Working | Marks | Guidance |
|---------|-------|---------|
| $x = 76°,\, 284°$ | B1 | 2 correct |
| $x = 109°,\, 251°$ (or better) | B1, B1 | Total: 3 marks. $-1$ each extra in range. If radians: $x = 1.32, 4.97, 1.91, 4.37$; B1 any 2 correct, B1 other 2 correct |
---4 It is given that $\tan ^ { 2 } x = \sec x + 11$.
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\tan ^ { 2 } x = \sec x + 11$ can be written in the form
$$\sec ^ { 2 } x - \sec x - 12 = 0$$
\item Hence show that $\cos x = \frac { 1 } { 4 }$ or $\cos x = - \frac { 1 } { 3 }$.
\item Hence, or otherwise, solve the equation $\tan ^ { 2 } x = \sec x + 11$, giving all values of $x$ to the nearest degree in the interval $0 ^ { \circ } < x < 360 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2005 Q4 [8]}}