| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2019 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | Multiple independent equations — includes show/prove component |
| Difficulty | Standard +0.3 Part (i) is a routine compound angle equation requiring basic inverse trig and CAST diagram work. Part (ii)(a) involves standard algebraic manipulation using tan θ = sin θ/cos θ and the Pythagorean identity—straightforward but requires careful algebra. Part (ii)(b) is solving a quadratic in cos θ, then finding angles. This is slightly above average due to the multi-step algebraic manipulation and the 'show that' proof component, but all techniques are standard C2 material with no novel insight required. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\sin(x + 60°) = -0.4\) → \(x + 60° = -23.6°\) → \(x = -83.6°\) | M1 A1 | Uses \(\arcsin(-0.4)\) to form one correct equation; either awrt \(-83.6°\) or \(143.6°\) |
| \(x + 60° = 203.6°\) → \(x = 143.6°\) | M1 A1 | Uses \(\arcsin(-0.4)\) to form both correct equations; both awrt \(-83.6°\) and \(143.6°\) with no additional solutions in range |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(2\sin\theta\tan\theta - 3 = \cos\theta\) → \(2\sin\theta\cdot\frac{\sin\theta}{\cos\theta} - 3 = \cos\theta\) | M1 | Uses \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) in given equation |
| \(\Rightarrow 2\sin^2\theta - 3\cos\theta = \cos^2\theta\) → \(\Rightarrow 2(1-\cos^2\theta) - 3\cos\theta = \cos^2\theta\) | M1 | Multiplies by \(\cos\theta\) and replaces \(\sin^2\theta\) by \(1 - \cos^2\theta\) to produce a 3TQ in \(\cos\theta\) |
| \(\Rightarrow 3\cos^2\theta + 3\cos\theta - 2 = 0\) | A1* | Correct completion; \(\cos\theta\) cannot appear as just \(\cos\); \(= 0\) must be seen on final line |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(3\cos^2\theta + 3\cos\theta - 2 = 0\) → \((\cos\theta) = \frac{-3 \pm \sqrt{33}}{6}\) | M1 | Attempts to solve by formula or completing the square ONLY; factorisation not allowed |
| \(\cos\theta = \frac{-3+\sqrt{33}}{6}\) or awrt \(\cos\theta = 0.46\) | A1 | Accept awrt \(0.46\); ignore reference to root \(< -1\) |
| \(\theta =\) awrt \(62.8°, 297.2°\) | dM1 A1 | dM1: uses arccos to reach at least one solution; A1: both answers awrt \(62.8°\) and awrt \(297.2°\) with no additional solutions in range |
## Question 14:
### Part 14(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $\sin(x + 60°) = -0.4$ → $x + 60° = -23.6°$ → $x = -83.6°$ | M1 A1 | Uses $\arcsin(-0.4)$ to form one correct equation; either awrt $-83.6°$ or $143.6°$ |
| $x + 60° = 203.6°$ → $x = 143.6°$ | M1 A1 | Uses $\arcsin(-0.4)$ to form both correct equations; both awrt $-83.6°$ and $143.6°$ with no additional solutions in range |
**Total: 4 marks**
---
### Part 14(ii)(a):
| Working | Mark | Guidance |
|---------|------|----------|
| $2\sin\theta\tan\theta - 3 = \cos\theta$ → $2\sin\theta\cdot\frac{\sin\theta}{\cos\theta} - 3 = \cos\theta$ | M1 | Uses $\tan\theta = \frac{\sin\theta}{\cos\theta}$ in given equation |
| $\Rightarrow 2\sin^2\theta - 3\cos\theta = \cos^2\theta$ → $\Rightarrow 2(1-\cos^2\theta) - 3\cos\theta = \cos^2\theta$ | M1 | Multiplies by $\cos\theta$ and replaces $\sin^2\theta$ by $1 - \cos^2\theta$ to produce a 3TQ in $\cos\theta$ |
| $\Rightarrow 3\cos^2\theta + 3\cos\theta - 2 = 0$ | A1* | Correct completion; $\cos\theta$ cannot appear as just $\cos$; $= 0$ must be seen on final line |
**Total: 3 marks**
---
### Part 14(ii)(b):
| Working | Mark | Guidance |
|---------|------|----------|
| $3\cos^2\theta + 3\cos\theta - 2 = 0$ → $(\cos\theta) = \frac{-3 \pm \sqrt{33}}{6}$ | M1 | Attempts to solve by formula or completing the square ONLY; factorisation not allowed |
| $\cos\theta = \frac{-3+\sqrt{33}}{6}$ or awrt $\cos\theta = 0.46$ | A1 | Accept awrt $0.46$; ignore reference to root $< -1$ |
| $\theta =$ awrt $62.8°, 297.2°$ | dM1 A1 | dM1: uses arccos to reach at least one solution; A1: both answers awrt $62.8°$ and awrt $297.2°$ with no additional solutions in range |
**Total: 4 marks**
---14. In this question solutions based entirely on graphical or numerical methods are not acceptable.
\begin{enumerate}[label=(\roman*)]
\item Solve, for $- 180 ^ { \circ } \leqslant x < 180 ^ { \circ }$, the equation
$$\sin \left( x + 60 ^ { \circ } \right) = - 0.4$$
giving your answers, in degrees, to one decimal place.
\item (a) Show that the equation
$$2 \sin \theta \tan \theta - 3 = \cos \theta$$
can be written in the form
$$3 \cos ^ { 2 } \theta + 3 \cos \theta - 2 = 0$$
(b) Hence solve, for $0 \leqslant \theta < 360 ^ { \circ }$, the equation
$$2 \sin \theta \tan \theta - 3 = \cos \theta$$
showing each stage of your working and giving your answers, in degrees, to one decimal place.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2019 Q14 [11]}}