| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Convert equation to quadratic form |
| Difficulty | Standard +0.3 Part (a) is a straightforward algebraic manipulation using tan²x = sin²x/cos²x and sin²x = 1-cos²x to reach the given quadratic form. Part (b) applies the same result to solve for θ in a triple angle equation, requiring careful consideration of the range and multiple solutions. This is slightly easier than average as it's a guided question with standard techniques, though the triple angle and range consideration add minor complexity. |
| 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 |
|---|---|---|
| \(2\cos x \dfrac{\sin^2 x}{\cos^2 x} = 3(1 + \cos x)\) | M1 (AO 3.1a) | Replaces \(\tan^2 x\) with \(\dfrac{\sin^2 x}{\cos^2 x}\) |
| \(2\cos x \left(\dfrac{1 - \cos^2 x}{\cos^2 x}\right) = 3(1 + \cos x)\) | M1 (AO 3.1a) | Replaces \(\sin^2 x\) with \(1 - \cos^2 x\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(5\cos^2 x + 3\cos x - 2 = 0\) | A1 (AO 2.1) | AG – correct working throughout; must show enough working to justify the given answer |
| Answer | Marks | Guidance |
|---|---|---|
| \((5\cos 3\theta - 2)(\cos 3\theta + 1) = 0\) | M1 (AO 1.1a) | Attempt to solve 3-term quadratic |
| \(\cos 3\theta = -1\) and \(\cos 3\theta = \dfrac{2}{5}\) | A1 (AO 2.1) | May be implied |
| \(\theta = \dfrac{1}{3}\arccos(-1)\), \(\theta = \dfrac{1}{3}\arccos\!\left(\dfrac{2}{5}\right)\) | M1 (AO 1.1) | Correct order of operation to find one value of \(\theta\) (or all values of \(3\theta\) correct); \((3\theta =)\ 66.42\ldots\), \(180\), \(293.57\ldots\) |
| \(60\) | A1 (AO 1.1) | Correct value (to at least 1 dp) |
| \(22.1\) | A1 (AO 1.1) | Correct value (to at least 1 dp) |
| \(97.9\) | A1 (AO 1.1) | Correct value (to at least 1 dp); any additional values in the range loses final A mark if earned |
## Question 5(a):
$2\cos x \tan^2 x = 3(1 + \cos x)$
$2\cos x \dfrac{\sin^2 x}{\cos^2 x} = 3(1 + \cos x)$ | M1 (AO 3.1a) | Replaces $\tan^2 x$ with $\dfrac{\sin^2 x}{\cos^2 x}$
$2\cos x \left(\dfrac{1 - \cos^2 x}{\cos^2 x}\right) = 3(1 + \cos x)$ | M1 (AO 3.1a) | Replaces $\sin^2 x$ with $1 - \cos^2 x$
$2(1 - \cos^2 x) = 3\cos x(1 + \cos x)$
$2 - 2\cos^2 x = 3\cos x + 3\cos^2 x$
$5\cos^2 x + 3\cos x - 2 = 0$ | A1 (AO 2.1) | AG – correct working throughout; must show enough working to justify the given answer
**Total: [3]**
## Question 5(b):
DR
$(5\cos 3\theta - 2)(\cos 3\theta + 1) = 0$ | M1 (AO 1.1a) | Attempt to solve 3-term quadratic
$\cos 3\theta = -1$ and $\cos 3\theta = \dfrac{2}{5}$ | A1 (AO 2.1) | May be implied
$\theta = \dfrac{1}{3}\arccos(-1)$, $\theta = \dfrac{1}{3}\arccos\!\left(\dfrac{2}{5}\right)$ | M1 (AO 1.1) | Correct order of operation to find one value of $\theta$ (or all values of $3\theta$ correct); $(3\theta =)\ 66.42\ldots$, $180$, $293.57\ldots$
$60$ | A1 (AO 1.1) | Correct value (to at least 1 dp)
$22.1$ | A1 (AO 1.1) | Correct value (to at least 1 dp)
$97.9$ | A1 (AO 1.1) | Correct value (to at least 1 dp); any additional values in the range loses final A mark if earned
**Total: [6]**
---5 (a) Show that the equation $2 \cos x \tan ^ { 2 } x = 3 ( 1 + \cos x )$ can be expressed in the form
$$5 \cos ^ { 2 } x + 3 \cos x - 2 = 0$$
\section*{(b) In this question you must show detailed reasoning.}
Hence solve the equation
$$2 \cos 3 \theta \tan ^ { 2 } 3 \theta = 3 ( 1 + \cos 3 \theta ) ,$$
giving all values of $\theta$ between $0 ^ { \circ }$ and $120 ^ { \circ }$, correct to $\mathbf { 1 }$ decimal place where appropriate.
\hfill \mbox{\textit{OCR PURE Q5 [9]}}