## Question 7:

### Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $2(\cos x°\cos 30° - \sin x°\sin 30°) = \sin x°\cos 30° - \cos x°\sin 30°$ | M1A1 | Correct expansion of both compound angle formulae |
| $\sin 30° = \frac{1}{2},\quad \cos 30° = \frac{\sqrt{3}}{2}$ | B1 | Correct values stated or used |
| $\tan x° = \frac{2\sqrt{3}+1}{\sqrt{3}+2} \times \frac{\sqrt{3}-2}{\sqrt{3}-2} \Rightarrow \tan x° = 3\sqrt{3}-4$ | dM1A1* | Rationalises denominator to reach given answer; correct solution only |

### Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan(2\theta+10)° = 3\sqrt{3}-4$ | M1 | Applies result from (a) with $x = 2\theta+10$ |
| $2\theta + 10 = 50.1,\ (230.1) \Rightarrow \theta = \ldots$ | dM1 | Solves for $\theta$ from correct equation |
| $\theta = 20.1,\quad 110.1$ | A1, A1 | Each correct value |

# Question (a) - Trigonometry:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses identities for $\cos(A+B)$ and $\sin(A-B)$ with $A=x, B=30$ | M1 | Condone missing bracket and incorrect signs but terms must be correct |
| Fully correct equation in $\sin x$ and $\cos x$ | A1 | |
| Replaces $\sin 30$ by $\frac{1}{2}$ and $\cos 30$ by $\frac{\sqrt{3}}{2}$ throughout expanded equation | B1 | If candidate divides by $\cos 30$ it will be for $\tan 30 = \frac{\sqrt{3}}{3}$ or equivalent |
| Collecting terms in $\sin x$ and $\cos x$ to reach $(\ldots)\sin x = (\ldots)\cos x$, then dividing by $\cos x$ to reach $\frac{\sin x}{\cos x} = \frac{(\ldots)}{(\ldots)}$ or $\tan x = \frac{(\ldots)}{(\ldots)}$ | dM1 | Intermediate line must be seen: $(\sqrt{3}+2)\tan x° = 2\sqrt{3}+1$ is required; without it is dM0 |
| $\tan x° = \frac{2\sqrt{3}+1}{\sqrt{3}+2} \times \frac{\sqrt{3}-2}{\sqrt{3}-2} \Rightarrow \tan x° = 3\sqrt{3}-4$ | A1* | Reaches final answer by showing rationalisation with no errors |

# Question (b) - Trigonometry:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses part (a) to produce equation $\tan(2\theta \pm \alpha)° = 3\sqrt{3}-4$ | M1 | Condone $\alpha=0$ and $\theta$ being replaced by $x$ |
| $\tan(2\theta \pm \alpha)° = 3\sqrt{3}-4 \Rightarrow \theta = \frac{\text{invtan}(3\sqrt{3}-4) \pm \alpha}{2}$ | dM1 | Dependent on previous M; correct method to find one value of $\theta$ |
| One correct answer awrt 1dp: $\theta = 20.1$ or $110.1$ | A1 | |
| Both $\theta = 20.1$ and $110.1$ awrt 1dp and no other solutions in range | A1 | Ignore extra solutions outside given range |

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