| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Show equation reduces to tan form |
| Difficulty | Standard +0.3 This is a straightforward application of standard trigonometric identities. Part (i) requires dividing both sides by cos to get tan, then using the addition formula for tan—a routine manipulation. Part (ii) uses the standard tan double angle formula to solve a quadratic, which is a common textbook exercise. The 'show that' structure and surd form answer are typical A-level features, but no novel insight is required. |
| 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 intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin 2\theta \cos\tfrac{\pi}{4} + \sin\tfrac{\pi}{4}\cos 2\theta = 3\cos 2\theta \cos\tfrac{\pi}{4} - 3\sin 2\theta \sin\tfrac{\pi}{4}\) | M1 1.1 | Correct use of compound angle formulae at least once |
| \(4\sin 2\theta = 2\cos 2\theta\) | A1 1.1 | Not from incorrect working |
| \(2\dfrac{\sin 2\theta}{\cos 2\theta} = 1 \Rightarrow \tan 2\theta = \dfrac{1}{2}\) | A1 2.2a [3] | AG — at least one step of intermediate working seen |
| ALT: \(\tan\!\left(2\theta + \tfrac{\pi}{4}\right) = 3\) | B1 | — |
| \(\dfrac{\tan 2\theta + 1}{1 - \tan 2\theta} = 3 \Rightarrow \tan 2\theta + 1 = 3(1 - \tan 2\theta)\) | M1 | Correct use of compound angle formula for tan and removal of fraction |
| \(\tan 2\theta = \dfrac{1}{2}\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan 2\theta = \dfrac{1}{2} \Rightarrow \dfrac{2\tan\theta}{1-\tan^2\theta} = \dfrac{1}{2}\) | M1* 3.1a | Double angle formula for \(\tan 2\theta\); allow one sign slip in formula |
| \(\tan^2\theta + 4\tan\theta - 1 = 0\) | Dep*M1 1.1 | Rearranges correctly to form 3-term quadratic in tan |
| \(\tan\theta = -2 \pm \sqrt{5}\) | A1 1.1 | BC — one correct exact value |
| \(-2 + \sqrt{5} > 0\) so \(\tan\theta = -2 + \sqrt{5}\) gives acute angle | A1 2.3 | Explicit rejection and reason for rejection |
| \(\therefore \tan\theta = -2 - \sqrt{5}\) | A1 2.2a [5] | This value only |
## Question 6(i):
$\sin\!\left(2\theta + \tfrac{\pi}{4}\right) = 3\cos\!\left(2\theta + \tfrac{\pi}{4}\right)$
$\sin 2\theta \cos\tfrac{\pi}{4} + \sin\tfrac{\pi}{4}\cos 2\theta = 3\cos 2\theta \cos\tfrac{\pi}{4} - 3\sin 2\theta \sin\tfrac{\pi}{4}$ | M1 1.1 | Correct use of compound angle formulae at least once
$4\sin 2\theta = 2\cos 2\theta$ | A1 1.1 | Not from incorrect working
$2\dfrac{\sin 2\theta}{\cos 2\theta} = 1 \Rightarrow \tan 2\theta = \dfrac{1}{2}$ | A1 2.2a [3] | **AG** — at least one step of intermediate working seen
**ALT:** $\tan\!\left(2\theta + \tfrac{\pi}{4}\right) = 3$ | B1 | —
$\dfrac{\tan 2\theta + 1}{1 - \tan 2\theta} = 3 \Rightarrow \tan 2\theta + 1 = 3(1 - \tan 2\theta)$ | M1 | Correct use of compound angle formula for tan and removal of fraction
$\tan 2\theta = \dfrac{1}{2}$ | A1 | —
---
## Question 6(ii):
$\tan 2\theta = \dfrac{1}{2} \Rightarrow \dfrac{2\tan\theta}{1-\tan^2\theta} = \dfrac{1}{2}$ | M1* 3.1a | Double angle formula for $\tan 2\theta$; allow one sign slip in formula
$\tan^2\theta + 4\tan\theta - 1 = 0$ | Dep*M1 1.1 | Rearranges correctly to form 3-term quadratic in tan
$\tan\theta = -2 \pm \sqrt{5}$ | A1 1.1 | **BC** — one correct exact value
$-2 + \sqrt{5} > 0$ so $\tan\theta = -2 + \sqrt{5}$ gives acute angle | A1 2.3 | Explicit rejection and reason for rejection
$\therefore \tan\theta = -2 - \sqrt{5}$ | A1 2.2a [5] | This value only
---6 It is given that the angle $\theta$ satisfies the equation $\sin \left( 2 \theta + \frac { 1 } { 4 } \pi \right) = 3 \cos \left( 2 \theta + \frac { 1 } { 4 } \pi \right)$.\\
(i) Show that $\tan 2 \theta = \frac { 1 } { 2 }$.\\
(ii) Hence find, in surd form, the exact value of $\tan \theta$, given that $\theta$ is an obtuse angle.
\hfill \mbox{\textit{OCR H240/03 2018 Q6 [8]}}