| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Expand compound angle then solve |
| Difficulty | Standard +0.3 This is a standard compound angle formula question requiring systematic expansion of sin(x-60°) and cos(x-30°), followed by algebraic manipulation to reach tan x. Part (b) uses substitution (2θ = x) to apply part (a), then solves for θ in the given range. While multi-step, it follows predictable patterns with no novel insight required, making it slightly easier than average. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts to use both \(\sin(x-60°) = \pm\sin x\cos 60° \pm \cos x\sin 60°\) and \(\cos(x-30°) = \pm\cos x\cos 30° \pm \sin x\sin 30°\) | M1 | 2.1 - Terms must be correct but condone sign errors and slip on multiplication of 2 |
| \(2\sin x\cos 60° - 2\cos x\sin 60° = \cos x\cos 30° + \sin x\sin 30°\) | A1 | 1.1b - Note \(\cos 60° = \sin 30°\) and \(\cos 30° = \sin 60°\). Also allow early substitution of trig values |
| Either uses \(\frac{\sin x}{\cos x} = \tan x\) to make \(\tan x\) subject, e.g. \((2\cos 60° - \sin 30°)\tan x = \cos 30° + 2\sin 60°\), OR uses exact values for \(\sin 30°\) etc with at least two correct, collects terms in \(\sin x\) and \(\cos x\), e.g. \(\left(2\times\frac{1}{2}-\frac{1}{2}\right)\sin x = \left(2\times\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right)\cos x\) | M1 | 2.1 - Must show: uses \(\frac{\sin x}{\cos x}=\tan x\); uses exact values with at least two correct; collects terms in \(\sin x\) and \(\cos x\) (two of three required) |
| \(\frac{1}{2}\tan x = \frac{3\sqrt{3}}{2} \Rightarrow \tan x = 3\sqrt{3}\) | A1* | 1.1b - Proceeds to given answer showing all key steps |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Deduces \(x = 2\theta + 60°\), e.g. \(\theta = \frac{x-60°}{2}\) | B1 | 2.2a - Implied by sight of \(\tan(2\theta+60°)=3\sqrt{3}\) |
| \(\tan(2\theta+60°) = 3\sqrt{3} \Rightarrow 2\theta+60° = 79.1°, 259.1°, \ldots\) | M1 | 1.1b - Proceeds from \(\tan(2\theta \pm \alpha°) = 3\sqrt{3}\); one angle for \(\arctan(3\sqrt{3})\) must be correct in degrees (FYI radian answers 1.38, 4.52) |
| Correct method to find one value of \(\theta\), e.g. \(\theta = \frac{79.1°-60°}{2}\) | dM1 | 1.1b - Dependent on one angle being correct in degrees for \(\arctan(3\sqrt{3})\) |
| \(\theta =\) awrt \(9.6°, 99.6°\) with no other values in range | A1 | 2.1 |
# Question 14:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to use both $\sin(x-60°) = \pm\sin x\cos 60° \pm \cos x\sin 60°$ and $\cos(x-30°) = \pm\cos x\cos 30° \pm \sin x\sin 30°$ | M1 | 2.1 - Terms must be correct but condone sign errors and slip on multiplication of 2 |
| $2\sin x\cos 60° - 2\cos x\sin 60° = \cos x\cos 30° + \sin x\sin 30°$ | A1 | 1.1b - Note $\cos 60° = \sin 30°$ and $\cos 30° = \sin 60°$. Also allow early substitution of trig values |
| Either uses $\frac{\sin x}{\cos x} = \tan x$ to make $\tan x$ subject, e.g. $(2\cos 60° - \sin 30°)\tan x = \cos 30° + 2\sin 60°$, OR uses exact values for $\sin 30°$ etc with at least two correct, collects terms in $\sin x$ and $\cos x$, e.g. $\left(2\times\frac{1}{2}-\frac{1}{2}\right)\sin x = \left(2\times\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right)\cos x$ | M1 | 2.1 - Must show: uses $\frac{\sin x}{\cos x}=\tan x$; uses exact values with at least two correct; collects terms in $\sin x$ and $\cos x$ (two of three required) |
| $\frac{1}{2}\tan x = \frac{3\sqrt{3}}{2} \Rightarrow \tan x = 3\sqrt{3}$ | A1* | 1.1b - Proceeds to given answer showing all key steps |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Deduces $x = 2\theta + 60°$, e.g. $\theta = \frac{x-60°}{2}$ | B1 | 2.2a - Implied by sight of $\tan(2\theta+60°)=3\sqrt{3}$ |
| $\tan(2\theta+60°) = 3\sqrt{3} \Rightarrow 2\theta+60° = 79.1°, 259.1°, \ldots$ | M1 | 1.1b - Proceeds from $\tan(2\theta \pm \alpha°) = 3\sqrt{3}$; one angle for $\arctan(3\sqrt{3})$ must be correct in degrees (FYI radian answers 1.38, 4.52) |
| Correct method to find one value of $\theta$, e.g. $\theta = \frac{79.1°-60°}{2}$ | dM1 | 1.1b - Dependent on one angle being correct in degrees for $\arctan(3\sqrt{3})$ |
| $\theta =$ awrt $9.6°, 99.6°$ with no other values in range | A1 | 2.1 |
**Alternative via** $\cos(2\theta+30°) = \cos 2\theta\cos 30° - \sin 2\theta\sin 30°$:
$2\sin 2\theta = \cos(2\theta+30°) \Rightarrow \tan 2\theta = \frac{\sqrt{3}}{5} \Rightarrow \theta = 9.6°, 99.6°$
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
Solutions relying entirely on calculator technology are not acceptable.\\
(a) Given that
$$2 \sin \left( x - 60 ^ { \circ } \right) = \cos \left( x - 30 ^ { \circ } \right)$$
show that
$$\tan x = 3 \sqrt { 3 }$$
(b) Hence or otherwise solve, for $0 \leqslant \theta < 180 ^ { \circ }$
$$2 \sin 2 \theta = \cos \left( 2 \theta + 30 ^ { \circ } \right)$$
giving your answers to one decimal place.
\hfill \mbox{\textit{Edexcel Paper 1 2022 Q14 [8]}}