| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.3 Part (a) is a straightforward identity proof using standard double angle formulae (cos 2θ = 1 - 2sin²θ and sin 2θ = 2sinθcosθ) with routine algebraic manipulation. Part (b) applies the proven identity to simplify an equation, then requires solving a quadratic in sec²x, which is a standard technique. The question involves multiple steps but uses well-practiced methods without requiring novel insight, making it slightly easier than average. |
| 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^21.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals1.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\tan\theta\sin 2\theta = \left(\frac{\sin\theta}{\cos\theta}\right)(2\sin\theta\cos\theta)\) | M1 | 1.1b — Applies \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) and \(\sin 2\theta = 2\sin\theta\cos\theta\) |
| \(= \left(\frac{\sin\theta}{\cos\theta}\right)(2\sin\theta\cos\theta) = 2\sin^2\theta = 1 - \cos 2\theta\) | M1, A1* | Cancels and uses \(\cos 2\theta = 1 - 2\sin^2\theta\); correct proof showing all steps |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(1 - \cos 2\theta = 1-(1-2\sin^2\theta) = 2\sin^2\theta\) | M1 | 1.1b — For using \(\cos 2\theta = 1-2\sin^2\theta\) |
| \(= \left(\frac{\sin\theta}{\cos\theta}\right)(2\sin\theta\cos\theta) = \tan\theta\sin 2\theta\) | M1, A1* | Correct proof showing all steps |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Deduces \(x = 0\) | B1 | 2.2a |
| Uses \(\sec^2 x = 1+\tan^2 x\) and cancels/factorises out \(\tan x\) or \((1-\cos 2x)\); e.g. \((1+\tan^2 x - 3\tan x - 5)\tan x = 0\) | M1 | 2.1 |
| \(\tan^2 x - 3\tan x - 4 = 0\) | A1 | 1.1b |
| \((\tan x-4)(\tan x+1) = 0 \Rightarrow \tan x = \ldots\) | M1 | 1.1b |
| \(x = -\frac{\pi}{4}, 1.326\) | A1, A1 | 1.1b — Only these two values in range \(-\frac{\pi}{2} < x < \frac{\pi}{2}\) |
| (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Deduces \(x=0\) | B1 | 2.2a |
| \(\sec^2 x - 5 = 3\tan x \Rightarrow \frac{1}{\cos^2 x}-5 = 3\left(\frac{\sin x}{\cos x}\right)\); \(1-5\cos^2 x = 3\sin x\cos x\); \(1-5\left(\frac{1+\cos 2x}{2}\right) = \frac{3}{2}\sin 2x\) | M1 | 2.1 — Complete process using identities for \(\sin 2x\) and \(\cos 2x\) |
| \(-\frac{3}{2} - \frac{5}{2}\cos 2x = \frac{3}{2}\sin 2x\) | A1 | 1.1b |
| \(\sqrt{34}\sin(2x+1.03) = -3\) | M1 | 1.1b — Expresses in form \(R\sin(2x+\alpha)=k\); \(k\neq 0\) |
| \(x = -\frac{\pi}{4}, 1.326\) | A1, A1 | 1.1b |
# Question 12:
## Part (a) Way 1:
| Working | Mark | Guidance |
|---------|------|----------|
| $\tan\theta\sin 2\theta = \left(\frac{\sin\theta}{\cos\theta}\right)(2\sin\theta\cos\theta)$ | M1 | 1.1b — Applies $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and $\sin 2\theta = 2\sin\theta\cos\theta$ |
| $= \left(\frac{\sin\theta}{\cos\theta}\right)(2\sin\theta\cos\theta) = 2\sin^2\theta = 1 - \cos 2\theta$ | M1, A1* | Cancels and uses $\cos 2\theta = 1 - 2\sin^2\theta$; correct proof showing all steps |
| | **(3)** | |
## Part (a) Way 2:
| Working | Mark | Guidance |
|---------|------|----------|
| $1 - \cos 2\theta = 1-(1-2\sin^2\theta) = 2\sin^2\theta$ | M1 | 1.1b — For using $\cos 2\theta = 1-2\sin^2\theta$ |
| $= \left(\frac{\sin\theta}{\cos\theta}\right)(2\sin\theta\cos\theta) = \tan\theta\sin 2\theta$ | M1, A1* | Correct proof showing all steps |
| | **(3)** | |
**Notes:**
- If $\cos 2\theta = \cos^2\theta - \sin^2\theta$ or $\cos 2\theta = 2\cos^2\theta - 1$ is used, mark cannot be awarded until $\cos^2\theta$ replaced by $1-\sin^2\theta$
- If proof meets in the middle, indication must be given that proof is complete
## Part (b) Way 1:
| Working | Mark | Guidance |
|---------|------|----------|
| Deduces $x = 0$ | B1 | 2.2a |
| Uses $\sec^2 x = 1+\tan^2 x$ and cancels/factorises out $\tan x$ or $(1-\cos 2x)$; e.g. $(1+\tan^2 x - 3\tan x - 5)\tan x = 0$ | M1 | 2.1 |
| $\tan^2 x - 3\tan x - 4 = 0$ | A1 | 1.1b |
| $(\tan x-4)(\tan x+1) = 0 \Rightarrow \tan x = \ldots$ | M1 | 1.1b |
| $x = -\frac{\pi}{4}, 1.326$ | A1, A1 | 1.1b — Only these two values in range $-\frac{\pi}{2} < x < \frac{\pi}{2}$ |
| | **(6)** | |
**Notes:**
- Allow $\pm\sec^2 x = \pm 1 \pm \tan^2 x$ for M1
- Correct 3TQ in $\tan x$: e.g. $\tan^2 x - 3\tan x - 4 = 0$
- A1: any one of $-\frac{\pi}{4}$, awrt $-0.785$, awrt $1.326$, $-45°$, awrt $75.964°$
- Final A1: only $x = -\frac{\pi}{4}, 1.326$ **cao** stated in range
## Part (b) Alternative Method 1:
| Working | Mark | Guidance |
|---------|------|----------|
| Deduces $x=0$ | B1 | 2.2a |
| $\sec^2 x - 5 = 3\tan x \Rightarrow \frac{1}{\cos^2 x}-5 = 3\left(\frac{\sin x}{\cos x}\right)$; $1-5\cos^2 x = 3\sin x\cos x$; $1-5\left(\frac{1+\cos 2x}{2}\right) = \frac{3}{2}\sin 2x$ | M1 | 2.1 — Complete process using identities for $\sin 2x$ and $\cos 2x$ |
| $-\frac{3}{2} - \frac{5}{2}\cos 2x = \frac{3}{2}\sin 2x$ | A1 | 1.1b |
| $\sqrt{34}\sin(2x+1.03) = -3$ | M1 | 1.1b — Expresses in form $R\sin(2x+\alpha)=k$; $k\neq 0$ |
| $x = -\frac{\pi}{4}, 1.326$ | A1, A1 | 1.1b |
---\begin{enumerate}
\item (a) Prove that
\end{enumerate}
$$1 - \cos 2 \theta \equiv \tan \theta \sin 2 \theta , \quad \theta \neq \frac { ( 2 n + 1 ) \pi } { 2 } , \quad n \in \mathbb { Z }$$
(b) Hence solve, for $- \frac { \pi } { 2 } < x < \frac { \pi } { 2 }$, the equation
$$\left( \sec ^ { 2 } x - 5 \right) ( 1 - \cos 2 x ) = 3 \tan ^ { 2 } x \sin 2 x$$
Give any non-exact answer to 3 decimal places where appropriate.
\hfill \mbox{\textit{Edexcel Paper 2 2018 Q12 [9]}}