| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Prove trig identity then solve |
| Difficulty | Standard +0.3 This is a standard C2 trigonometric equation requiring algebraic manipulation using tan x = sin x/cos x, followed by solving a quadratic in cos x and finding angles. The 'show that' part guides students through the manipulation, making it slightly easier than average but still requiring multiple steps and careful algebra. |
| 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 | Marks | Guidance |
| \(2\sin x \cdot \frac{\sin x}{\cos x} = 4\cos x - 1\), \(2\sin^2 x = 4\cos^2 x - \cos x\) | M1 | Use \(\tan x = \frac{\sin x}{\cos x}\) and rearrange to a form not involving fractions. Must multiply all terms by \(\cos x\) so \(4\cos^2 x - 1\) is M0, but allow M1 for \(\cos x(4\cos x - 1)\) even if subsequent errors |
| \(2 - 2\cos^2 x = 4\cos^2 x - \cos x\) | M1 | Use \(\sin^2 x = 1 - \cos^2 x\). Must be used correctly, so M0 for \(1 - 2\cos^2 x\). Must be attempting quadratic in \(\cos x\) so M0 for \(\cos^2 x = 1 - \sin^2 x\) |
| \(6\cos^2 x - \cos x - 2 = 0\) | A1 | Must be equation ie \(= 0\). Allow poor notation as long as final answer is correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((3\cos x - 2)(2\cos x + 1) = 0\) | M1 | Attempt to solve quadratic in \(\cos x\). This mark is just for solving a 3 term quadratic. Condone any substitution used, including \(x = \cos x\) |
| \(\cos x = \frac{2}{3}\), \(\cos x = -\frac{1}{2}\) | M1 | Attempt to find \(x\) from root(s) of quadratic. Attempt \(\cos^{-1}\) of at least one root. Allow for just stating \(\cos^{-1}\) (their root) inc if \( |
| At least 2 correct angles | A1 | Allow 3sf or better. Must come from correct solution of quadratic. Allow radian equivs \(0.841\), \(5.44\), \(\frac{2\pi}{3}\) or \(2.09\), \(\frac{4\pi}{3}\) or \(4.19\) |
| \(x = 48.2°, 312°, 120°, 240°\) all correct, no extras | A1 | Must be in degrees. SR if no working shown then allow B1 for 2 correct angles or B2 for 4 correct angles, no extras |
## Question 5:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2\sin x \cdot \frac{\sin x}{\cos x} = 4\cos x - 1$, $2\sin^2 x = 4\cos^2 x - \cos x$ | M1 | Use $\tan x = \frac{\sin x}{\cos x}$ and rearrange to a form not involving fractions. Must multiply all terms by $\cos x$ so $4\cos^2 x - 1$ is M0, but allow M1 for $\cos x(4\cos x - 1)$ even if subsequent errors |
| $2 - 2\cos^2 x = 4\cos^2 x - \cos x$ | M1 | Use $\sin^2 x = 1 - \cos^2 x$. Must be used correctly, so M0 for $1 - 2\cos^2 x$. Must be attempting quadratic in $\cos x$ so M0 for $\cos^2 x = 1 - \sin^2 x$ |
| $6\cos^2 x - \cos x - 2 = 0$ | A1 | Must be equation ie $= 0$. Allow poor notation as long as final answer is correct |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(3\cos x - 2)(2\cos x + 1) = 0$ | M1 | Attempt to solve quadratic in $\cos x$. This mark is just for solving a 3 term quadratic. Condone any substitution used, including $x = \cos x$ |
| $\cos x = \frac{2}{3}$, $\cos x = -\frac{1}{2}$ | M1 | Attempt to find $x$ from root(s) of quadratic. Attempt $\cos^{-1}$ of at least one root. Allow for just stating $\cos^{-1}$ (their root) inc if $|\cos x| > 1$ |
| At least 2 correct angles | A1 | Allow 3sf or better. Must come from correct solution of quadratic. Allow radian equivs $0.841$, $5.44$, $\frac{2\pi}{3}$ or $2.09$, $\frac{4\pi}{3}$ or $4.19$ |
| $x = 48.2°, 312°, 120°, 240°$ all correct, no extras | A1 | Must be in degrees. **SR** if no working shown then allow **B1** for 2 correct angles or **B2** for 4 correct angles, no extras |
---5 (i) Show that the equation $2 \sin x = \frac { 4 \cos x - 1 } { \tan x }$ can be expressed in the form
$$6 \cos ^ { 2 } x - \cos x - 2 = 0 .$$
(ii) Hence solve the equation $2 \sin x = \frac { 4 \cos x - 1 } { \tan x }$, giving all values of $x$ between $0 ^ { \circ }$ and $360 ^ { \circ }$.
\hfill \mbox{\textit{OCR C2 2013 Q5 [7]}}