| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | September |
| Marks | 7 |
| Topic | Standard trigonometric equations |
| Type | Equation with 'show that' rewriting preliminary part |
| Difficulty | Standard +0.3 This is a straightforward multi-part trigonometric equation question requiring standard techniques: algebraic manipulation using sin²θ + cos²θ = 1, solving a quadratic in cos θ, and checking validity of solutions. While it has multiple parts and requires careful reasoning about extraneous solutions from squaring, these are routine A-level skills with no novel insight needed, making it slightly easier than average. |
| 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 |
|---|---|---|
| \(2(1 - \cos^2\theta) + \cos\theta = 4\cos^2\theta\) | M1 | Correctly removing square root and use of \(\sin^2\theta = 1 - \cos^2\theta\) to obtain an equation in \(\cos\) only |
| Answer | Marks | Guidance |
|---|---|---|
| \(6\cos^2\theta - \cos\theta - 2 = 0\) | A1 | AG – sufficient working must be shown to establish given result |
| Answer | Marks | Guidance |
|---|---|---|
| (DR) \((2\cos\theta + 1)(3\cos\theta - 2) = 0\) | M1 | Correct method for solving quadratic |
| \(\cos\theta = -\frac{1}{2}\) and \(\cos\theta = \frac{2}{3}\) | A1 | |
| \(\cos\theta = \frac{2}{3} \Rightarrow \theta = 48.2, 311.8\) | A1 | Any two correct values |
| \(\cos\theta = -\frac{1}{2} \Rightarrow \theta = 120, 240\) | A1 | All four correct values |
| Answer | Marks |
|---|---|
| E.g. \(\cos\theta \neq -\frac{1}{2}\) since \(\cos\theta \geq 0\) in the RHS of the equation \(\sqrt{2\sin^2\theta + \cos\theta} = 2\cos\theta\) | E1 |
### (i)
$2(1 - \cos^2\theta) + \cos\theta = 4\cos^2\theta$ | M1 | Correctly removing square root and use of $\sin^2\theta = 1 - \cos^2\theta$ to obtain an equation in $\cos$ only
$2 - 2\cos^2\theta + \cos\theta = 4\cos^2\theta$
$6\cos^2\theta - \cos\theta - 2 = 0$ | A1 | AG – sufficient working must be shown to establish given result
**[2 marks total]**
### (ii)
**(DR)** $(2\cos\theta + 1)(3\cos\theta - 2) = 0$ | M1 | Correct method for solving quadratic
$\cos\theta = -\frac{1}{2}$ and $\cos\theta = \frac{2}{3}$ | A1 |
$\cos\theta = \frac{2}{3} \Rightarrow \theta = 48.2, 311.8$ | A1 | Any two correct values
$\cos\theta = -\frac{1}{2} \Rightarrow \theta = 120, 240$ | A1 | All four correct values
**[4 marks total]**
### (iii)
E.g. $\cos\theta \neq -\frac{1}{2}$ since $\cos\theta \geq 0$ in the RHS of the equation $\sqrt{2\sin^2\theta + \cos\theta} = 2\cos\theta$ | E1 |
**[1 mark total]**
---3 (i) Given that $\sqrt { 2 \sin ^ { 2 } \theta + \cos \theta } = 2 \cos \theta$, show that $6 \cos ^ { 2 } \theta - \cos \theta - 2 = 0$.\\
(ii) In this question you must show detailed reasoning.
Solve the equation
$$6 \cos ^ { 2 } \theta - \cos \theta - 2 = 0$$
giving all values of $\theta$ between $0 ^ { \circ }$ and $360 ^ { \circ }$ correct to 1 decimal place.\\
(iii) Explain why not all the solutions from part (ii) are solutions of the equation
$$\sqrt { 2 \sin ^ { 2 } \theta + \cos \theta } = 2 \cos \theta$$
\hfill \mbox{\textit{OCR H240/03 2018 Q3 [7]}}