| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Quadratic trigonometric equations |
| Type | Show then solve: sin²/cos² substitution |
| Difficulty | Moderate -0.3 Part (a) is straightforward inverse trig with interval consideration. Part (b)(i) uses the standard identity sin²θ + cos²θ = 1 to form a quadratic in cos θ—a routine C2 technique. Part (b)(ii) requires finding tan θ from cos θ, which is standard but adds a step. Overall slightly easier than average due to predictable structure and standard methods. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x + 0.7 = \arcsin(0.6) = 0.6435...\) | M1 | Correct first solution for \(x + 0.7\) |
| \(x + 0.7 = \pi - 0.6435... = 2.498...\) | M1 | Second solution using \(\pi - \arcsin\) |
| \(x = -0.06\) (2 s.f.) | A1 | Both answers required |
| \(x = 1.8\) (2 s.f.) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use \(\sin^2\theta = 1 - \cos^2\theta\) | M1 | Substituting identity |
| \(5\cos^2\theta - \cos\theta = 1 - \cos^2\theta\) | ||
| \(6\cos^2\theta - \cos\theta - 1 = 0\) | A1 | Correct quadratic |
| \((3\cos\theta + 1)(2\cos\theta - 1) = 0\) | M1 | Attempt to solve quadratic |
| \(\cos\theta = \frac{1}{2}\) or \(\cos\theta = -\frac{1}{3}\) | A1 | Both values correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(\cos\theta = -\frac{1}{3}\), use \(\sin^2\theta = 1 - \frac{1}{9} = \frac{8}{9}\) | M1 | Using Pythagoras/identity with \(\cos\theta = -\frac{1}{3}\) |
| \(\sin\theta = \pm\frac{2\sqrt{2}}{3}\) | A1 | Correct value of \(\sin\theta\) |
| \(\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\pm\frac{2\sqrt{2}}{3}}{-\frac{1}{3}} = \mp 2\sqrt{2}\) | A1 | Correct completion showing \(\tan\theta = 2\sqrt{2}\) (taking appropriate sign) |
## Question 6:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x + 0.7 = \arcsin(0.6) = 0.6435...$ | M1 | Correct first solution for $x + 0.7$ |
| $x + 0.7 = \pi - 0.6435... = 2.498...$ | M1 | Second solution using $\pi - \arcsin$ |
| $x = -0.06$ (2 s.f.) | A1 | Both answers required |
| $x = 1.8$ (2 s.f.) | | |
### Part (b)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $\sin^2\theta = 1 - \cos^2\theta$ | M1 | Substituting identity |
| $5\cos^2\theta - \cos\theta = 1 - \cos^2\theta$ | | |
| $6\cos^2\theta - \cos\theta - 1 = 0$ | A1 | Correct quadratic |
| $(3\cos\theta + 1)(2\cos\theta - 1) = 0$ | M1 | Attempt to solve quadratic |
| $\cos\theta = \frac{1}{2}$ or $\cos\theta = -\frac{1}{3}$ | A1 | Both values correct |
### Part (b)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $\cos\theta = -\frac{1}{3}$, use $\sin^2\theta = 1 - \frac{1}{9} = \frac{8}{9}$ | M1 | Using Pythagoras/identity with $\cos\theta = -\frac{1}{3}$ |
| $\sin\theta = \pm\frac{2\sqrt{2}}{3}$ | A1 | Correct value of $\sin\theta$ |
| $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\pm\frac{2\sqrt{2}}{3}}{-\frac{1}{3}} = \mp 2\sqrt{2}$ | A1 | Correct completion showing $\tan\theta = 2\sqrt{2}$ (taking appropriate sign) |
---6
\begin{enumerate}[label=(\alph*)]
\item Solve the equation $\sin ( x + 0.7 ) = 0.6$ in the interval $- \pi < x < \pi$, giving your answers in radians to two significant figures.
\item It is given that $5 \cos ^ { 2 } \theta - \cos \theta = \sin ^ { 2 } \theta$.
\begin{enumerate}[label=(\roman*)]
\item By forming and solving a suitable quadratic equation, find the possible values of $\cos \theta$.
\item Hence show that a possible value of $\tan \theta$ is $2 \sqrt { 2 }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2015 Q6 [10]}}