| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | January |
| 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 standard C3 reciprocal trig question requiring the identity cot²x + 1 = cosec²x, solving a quadratic in cosec x, and applying a compound angle shift. All steps are routine textbook techniques with no novel insight required, 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2(\csc^2 x - 1) + 5\csc x = 10\) | M1 | |
| \(2\csc^2 x - 2 + 5\csc x - 10 = 0\) | ||
| \(2\csc^2 x + 5\csc x - 12 = 0\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((2\csc x - 3)(\csc x + 4) = 0\) | M1 | Attempt to solve |
| \(\csc x = \dfrac{3}{2}\) or \(-4\) | A1 | Condone answers with no method shown |
| \(\sin x = \dfrac{2}{3}\) or \(-\dfrac{1}{4}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((\theta - 0.1) = 0.73, 2.41, -0.25, -2.89\) | B1 | 2 correct values, may be implied later; \((41.8°, 138.2°, -165.5°, -14.5°)\) |
| \(\theta = 0.83, 2.51, -0.15, -2.79\) (AWRT) | B1 | 2 correct answers |
| B1 | +2 correct answers and no extra within range |
## Question 5:
### Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2(\csc^2 x - 1) + 5\csc x = 10$ | M1 | |
| $2\csc^2 x - 2 + 5\csc x - 10 = 0$ | | |
| $2\csc^2 x + 5\csc x - 12 = 0$ | A1 | AG |
### Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(2\csc x - 3)(\csc x + 4) = 0$ | M1 | Attempt to solve |
| $\csc x = \dfrac{3}{2}$ or $-4$ | A1 | Condone answers with no method shown |
| $\sin x = \dfrac{2}{3}$ or $-\dfrac{1}{4}$ | A1 | AG |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(\theta - 0.1) = 0.73, 2.41, -0.25, -2.89$ | B1 | 2 correct values, may be implied later; $(41.8°, 138.2°, -165.5°, -14.5°)$ |
| $\theta = 0.83, 2.51, -0.15, -2.79$ (AWRT) | B1 | 2 correct answers |
| | B1 | +2 correct answers and no extra within range |
**Total: 8 marks**
---5
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the equation
$$2 \cot ^ { 2 } x + 5 \operatorname { cosec } x = 10$$
can be written in the form $2 \operatorname { cosec } ^ { 2 } x + 5 \operatorname { cosec } x - 12 = 0$.
\item Hence show that $\sin x = - \frac { 1 } { 4 }$ or $\sin x = \frac { 2 } { 3 }$.
\end{enumerate}\item Hence, or otherwise, solve the equation
$$2 \cot ^ { 2 } ( \theta - 0.1 ) + 5 \operatorname { cosec } ( \theta - 0.1 ) = 10$$
giving all values of $\theta$ in radians to two decimal places in the interval $- \pi < \theta < \pi$.\\
(3 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2007 Q5 [8]}}