| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2011 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Convert equation to quadratic form |
| Difficulty | Standard +0.3 This is a structured multi-part question with clear scaffolding. Part (a) is routine recall (sec x = -5 → cos x = -1/5). Part (b) involves algebraic manipulation of reciprocal trig functions following standard techniques (common denominator, Pythagorean identity). Part (c) directly uses the result from (b) and mirrors part (a). The algebraic manipulation requires care but follows predictable patterns taught in C3, making this 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=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\cos x = -\frac{1}{5}\) | M1 | Using \(\sec x = \frac{1}{\cos x}\) |
| \(x = \arccos(-\tfrac{1}{5}) = 1.77\) | A1 | One correct value |
| \(x = 2\pi - 1.77 = 4.51\) | A1 | Second correct value in interval |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{\csc x(1-\csc x) - \csc x(1+\csc x)}{(1+\csc x)(1-\csc x)} = 50\) | M1 | Combining fractions over common denominator |
| Numerator: \(\csc x - \csc^2 x - \csc x - \csc^2 x = -2\csc^2 x\) | A1 | Correct numerator |
| Denominator: \(1 - \csc^2 x = -\cot^2 x\) | A1 | Using identity correctly |
| \(\frac{-2\csc^2 x}{-\cot^2 x} = \frac{2\csc^2 x}{\cot^2 x} = 2\cdot\frac{1/\sin^2 x}{\cos^2 x/\sin^2 x} = \frac{2}{\cos^2 x} = 2\sec^2 x = 50\) | M1A1 | Reaching \(\sec^2 x = 25\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sec^2 x = 25 \Rightarrow \sec x = \pm 5\) | M1 | Using result from (b) |
| \(\sec x = -5\): solutions \(x = 1.77, 4.51\) from part (a) | A1 | Correct solutions from \(\sec x = -5\) |
| \(\sec x = 5 \Rightarrow \cos x = \frac{1}{5} \Rightarrow x = 1.37, 4.91\) | A1 | Both additional solutions |
# Question 7:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\cos x = -\frac{1}{5}$ | M1 | Using $\sec x = \frac{1}{\cos x}$ |
| $x = \arccos(-\tfrac{1}{5}) = 1.77$ | A1 | One correct value |
| $x = 2\pi - 1.77 = 4.51$ | A1 | Second correct value in interval |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{\csc x(1-\csc x) - \csc x(1+\csc x)}{(1+\csc x)(1-\csc x)} = 50$ | M1 | Combining fractions over common denominator |
| Numerator: $\csc x - \csc^2 x - \csc x - \csc^2 x = -2\csc^2 x$ | A1 | Correct numerator |
| Denominator: $1 - \csc^2 x = -\cot^2 x$ | A1 | Using identity correctly |
| $\frac{-2\csc^2 x}{-\cot^2 x} = \frac{2\csc^2 x}{\cot^2 x} = 2\cdot\frac{1/\sin^2 x}{\cos^2 x/\sin^2 x} = \frac{2}{\cos^2 x} = 2\sec^2 x = 50$ | M1A1 | Reaching $\sec^2 x = 25$ |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sec^2 x = 25 \Rightarrow \sec x = \pm 5$ | M1 | Using result from (b) |
| $\sec x = -5$: solutions $x = 1.77, 4.51$ from part (a) | A1 | Correct solutions from $\sec x = -5$ |
| $\sec x = 5 \Rightarrow \cos x = \frac{1}{5} \Rightarrow x = 1.37, 4.91$ | A1 | Both additional solutions |
---7
\begin{enumerate}[label=(\alph*)]
\item Solve the equation $\sec x = - 5$, giving all values of $x$ in radians to two decimal places in the interval $0 < x < 2 \pi$.
\item Show that the equation
$$\frac { \operatorname { cosec } x } { 1 + \operatorname { cosec } x } - \frac { \operatorname { cosec } x } { 1 - \operatorname { cosec } x } = 50$$
can be written in the form
$$\sec ^ { 2 } x = 25$$
\item Hence, or otherwise, solve the equation
$$\frac { \operatorname { cosec } x } { 1 + \operatorname { cosec } x } - \frac { \operatorname { cosec } x } { 1 - \operatorname { cosec } x } = 50$$
giving all values of $x$ in radians to two decimal places in the interval $0 < x < 2 \pi$.\\
(3 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2011 Q7 [10]}}