| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Show then solve substituted equation |
| Difficulty | Moderate -0.3 This is a straightforward C2 trigonometric equation requiring standard identity substitution (cos²x = 1-sin²x) to simplify to tan²x, then applying the result to a compound angle case. The algebraic manipulation is routine and the question structure is highly scaffolded with part (a) doing most of the work. Slightly easier than average due to the 'show that' guidance and standard techniques. |
| 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 | Mark | Guidance |
| Use \(\cos^2 x = 1 - \sin^2 x\) in numerator: \(\frac{1 - \sin^2 x + 4\sin^2 x}{1-\sin^2 x} = 7\) | M1 | Substituting identity |
| \(\frac{1 + 3\sin^2 x}{\cos^2 x} = 7\) | A1 | Correct simplification |
| \(1 + 3\sin^2 x = 7\cos^2 x = 7(1-\sin^2 x)\), leading to \(10\sin^2 x = 6\), so \(\tan^2 x = \frac{\sin^2 x}{\cos^2 x} = \frac{3}{2}\) | A1 | Correct completion, shown convincingly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\tan^2 2\theta = \frac{3}{2}\), so \(\tan 2\theta = \pm\sqrt{\frac{3}{2}}\) | M1 | Using result from (a) with \(2\theta\) |
| \(2\theta = \arctan\left(\sqrt{\frac{3}{2}}\right) = 50.77...°\) | A1 | One correct value |
| \(2\theta = 50.8°, 129.2°, 230.8°, 309.2°\) | A1 | All four values of \(2\theta\) |
| \(\theta = 25°, 65°, 115°, 155°\) | A1 | All four correct values of \(\theta\) to nearest degree |
# Question 7:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use $\cos^2 x = 1 - \sin^2 x$ in numerator: $\frac{1 - \sin^2 x + 4\sin^2 x}{1-\sin^2 x} = 7$ | M1 | Substituting identity |
| $\frac{1 + 3\sin^2 x}{\cos^2 x} = 7$ | A1 | Correct simplification |
| $1 + 3\sin^2 x = 7\cos^2 x = 7(1-\sin^2 x)$, leading to $10\sin^2 x = 6$, so $\tan^2 x = \frac{\sin^2 x}{\cos^2 x} = \frac{3}{2}$ | A1 | Correct completion, shown convincingly |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\tan^2 2\theta = \frac{3}{2}$, so $\tan 2\theta = \pm\sqrt{\frac{3}{2}}$ | M1 | Using result from (a) with $2\theta$ |
| $2\theta = \arctan\left(\sqrt{\frac{3}{2}}\right) = 50.77...°$ | A1 | One correct value |
| $2\theta = 50.8°, 129.2°, 230.8°, 309.2°$ | A1 | All four values of $2\theta$ |
| $\theta = 25°, 65°, 115°, 155°$ | A1 | All four correct values of $\theta$ to nearest degree |
---7
\begin{enumerate}[label=(\alph*)]
\item Given that $\frac { \cos ^ { 2 } x + 4 \sin ^ { 2 } x } { 1 - \sin ^ { 2 } x } = 7$, show that $\tan ^ { 2 } x = \frac { 3 } { 2 }$.
\item Hence solve the equation $\frac { \cos ^ { 2 } 2 \theta + 4 \sin ^ { 2 } 2 \theta } { 1 - \sin ^ { 2 } 2 \theta } = 7$ in the interval $0 ^ { \circ } < \theta < 180 ^ { \circ }$, giving your values of $\theta$ to the nearest degree.\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2014 Q7 [7]}}