| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2015 |
| Session | June |
| 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 straightforward application of standard reciprocal trig identities (cosec²θ = 1 + cot²θ) to convert to a quadratic form, followed by routine solving. Part (a) is algebraic manipulation requiring one key identity, and part (b) is standard substitution and solving for angles in a given range. Slightly easier than average due to the 'show that' structure guiding students through the method. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals |
## Question 8(a):
Show $4\cosec^2\theta - \cot^2\theta = k$ gives $\sec^2\theta = \frac{k-1}{k-4}$
**Working:**
- Use $\cosec^2\theta = 1 + \cot^2\theta$
- $4(1 + \cot^2\theta) - \cot^2\theta = k$
- $3\cot^2\theta = k - 4$, so $\cot^2\theta = \frac{k-4}{3}$
- $\tan^2\theta = \frac{3}{k-4}$
- $\sec^2\theta = 1 + \tan^2\theta = \frac{k-4+3}{k-4} = \frac{k-1}{k-4}$ ✓
## Question 8(b): with $k=5$, $\theta = 2x+75°$
- $\sec^2(2x+75°) = \frac{4}{1} = 4$
- $\cos^2(2x+75°) = \frac{1}{4}$
- $2x + 75° = 60°, 120°, 240°, 300°...$
- Valid solutions: $x = 22.5°, 52.5°$
For the **official mark scheme**, please consult AQA's website directly.8
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $4 \operatorname { cosec } ^ { 2 } \theta - \cot ^ { 2 } \theta = k$, where $k \neq 4$, can be written in the form
$$\sec ^ { 2 } \theta = \frac { k - 1 } { k - 4 }$$
\item Hence, or otherwise, solve the equation
$$4 \operatorname { cosec } ^ { 2 } \left( 2 x + 75 ^ { \circ } \right) - \cot ^ { 2 } \left( 2 x + 75 ^ { \circ } \right) = 5$$
giving all values of $x$ in the interval $0 ^ { \circ } < x < 180 ^ { \circ }$.\\[0pt]
[5 marks]\\
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\hfill \mbox{\textit{AQA C3 2015 Q8 [10]}}