| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Convert equation to quadratic form |
| Difficulty | Standard +0.3 This is a straightforward two-part question requiring standard double angle identities (cos 2x = 1 - 2sin²x, sin 2x = 2sin x cos x) for the proof, then using the result to convert a sec equation to tan, leading to a simple quadratic. All techniques are routine C3/C4 material with no novel insight required, making it slightly easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals1.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1-\cos 2x}{\sin 2x} \equiv \frac{1-(1-2\sin^2 x)}{2\sin x \cos x} = \frac{2\sin x \sin x}{2\sin x \cos x}\) | M1 A1 | M1: Uses \(\cos 2x = 1-2\sin^2 x\) and \(\sin 2x = 2\sin x\cos x\); A1: Correct intermediate line |
| \(= \frac{\sin x}{\cos x} = \tan x\) | A1* | Must show cancelling step; no errors or omissions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3\sec^2\theta - 7 = \frac{1-\cos 2\theta}{\sin 2\theta} \Rightarrow 3\sec^2\theta - 7 = \tan\theta\) | M1 | Uses identity from part (a) |
| \(\Rightarrow 3(1+\tan^2\theta) - 7 = \tan\theta\) | M1 | Uses \(\sec^2\theta = 1 \pm \tan^2\theta\) |
| \(\Rightarrow 3\tan^2\theta - \tan\theta - 4 = 0\) | A1 | Correct equation in \(\tan\theta\) |
| \(\Rightarrow (3\tan\theta - 4)(\tan\theta + 1) = 0\) | ||
| \(\Rightarrow \tan\theta = \frac{4}{3},\ \tan\theta = -1\) | dM1 | Dependent on both previous M marks |
| \(\theta = 0.927,\ 4.069,\ \frac{3}{4}\pi\ (2.356),\ \frac{7}{4}\pi\ (5.498)\) | A1 A1 |
# Question 4:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1-\cos 2x}{\sin 2x} \equiv \frac{1-(1-2\sin^2 x)}{2\sin x \cos x} = \frac{2\sin x \sin x}{2\sin x \cos x}$ | M1 A1 | M1: Uses $\cos 2x = 1-2\sin^2 x$ **and** $\sin 2x = 2\sin x\cos x$; A1: Correct intermediate line |
| $= \frac{\sin x}{\cos x} = \tan x$ | A1* | Must show cancelling step; no errors or omissions |
**(3 marks)**
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3\sec^2\theta - 7 = \frac{1-\cos 2\theta}{\sin 2\theta} \Rightarrow 3\sec^2\theta - 7 = \tan\theta$ | M1 | Uses identity from part (a) |
| $\Rightarrow 3(1+\tan^2\theta) - 7 = \tan\theta$ | M1 | Uses $\sec^2\theta = 1 \pm \tan^2\theta$ |
| $\Rightarrow 3\tan^2\theta - \tan\theta - 4 = 0$ | A1 | Correct equation in $\tan\theta$ |
| $\Rightarrow (3\tan\theta - 4)(\tan\theta + 1) = 0$ | | |
| $\Rightarrow \tan\theta = \frac{4}{3},\ \tan\theta = -1$ | dM1 | Dependent on both previous M marks |
| $\theta = 0.927,\ 4.069,\ \frac{3}{4}\pi\ (2.356),\ \frac{7}{4}\pi\ (5.498)$ | A1 A1 | |
**(6 marks, 9 marks total)**4. (a) Prove that
$$\frac { 1 - \cos 2 x } { \sin 2 x } \equiv \tan x , \quad x \neq \frac { n \pi } { 2 }$$
(b) Hence solve, for $0 \leqslant \theta < 2 \pi$,
$$3 \sec ^ { 2 } \theta - 7 = \frac { 1 - \cos 2 \theta } { \sin 2 \theta }$$
Give your answers in radians to 3 decimal places, as appropriate.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)\\
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\hfill \mbox{\textit{Edexcel C34 2017 Q4 [9]}}