# Question 1:

## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sin x = \frac{2t}{1+t^2}$ | B1 | AO1.2 |

## Part (b)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\tan\left(\frac{x}{2}\right) = \sqrt{2} \Rightarrow t = \sqrt{2} \Rightarrow \sin x = \frac{2(\sqrt{2})}{1+(\sqrt{2})^2}$ or $\frac{2(\sqrt{2})}{1+2}$ | M1 | Complete substitution of $t=\sqrt{2}$ into expression from part (a). AO1.1b |
| $\sin x = \frac{2}{3}\sqrt{2}$ or $\frac{1}{3}\sqrt{8}$ or $\sqrt{\frac{8}{9}}$ | A1 | Correct exact answer. AO1.1b. Note: Give M0A0 for correct answer without evidence of substitution. $\sin x = \frac{2}{3}\sqrt{2} = 0.9428...$ |

## Part (b)(ii) — Way 1
| Answer | Mark | Guidance |
|--------|------|----------|
| $\cos x = \frac{\sin x}{\tan x} \Rightarrow \cos x = \frac{\frac{2t}{1+t^2}}{\frac{2t}{1-t^2}}$ | M1 | Uses correct trig identity to find expression connecting only $\cos x$ (or $\cos^2 x$) and $t$. AO1.1b |
| $\cos x = \frac{1-t^2}{1+t^2}$ * cso | A1* | Correct proof. AO2.1 |

## Part (b)(ii) — Way 2
| Answer | Mark | Guidance |
|--------|------|----------|
| $\tan x = \frac{\sin x}{\cos x} \Rightarrow \frac{\frac{2t}{1+t^2}}{\frac{1-t^2}{}} = \frac{1}{\cos x} \Rightarrow \cos x = \frac{1-t^2}{1+t^2}$ * cso | M1, A1* | AO1.1b, AO2.1 |

## Part (b)(ii) — Way 3
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sin^2 x + \cos^2 x \equiv 1 \Rightarrow \left(\frac{2t}{1+t^2}\right)^2 + \cos^2 x = 1$ | M1 | AO1.1b |
| $\cos^2 x = 1 - \left(\frac{2t}{1+t^2}\right)^2 = \frac{(1+t^2)^2 - 4t^2}{(1+t^2)^2} = \frac{1-2t^2+t^4}{(1+t^2)^2} = \frac{(1-t^2)^2}{(1+t^2)^2}$ $\Rightarrow \cos x = \frac{1-t^2}{1+t^2}$ * cso | A1 | AO2.1 |

## Part (b)(ii) — Way 4
| Answer | Mark | Guidance |
|--------|------|----------|
| $o^2 + a^2 = h^2 \Rightarrow (2t)^2 + a^2 = (1+t^2)^2$ | M1 | Uses $\sin x = \frac{o}{h}$ and correct Pythagoras to express adjacent edge in terms of $t$. AO1.1b |
| $a^2 = (1+t^2)^2 - (2t)^2 = 1-2t^2+t^4 = (1-t^2)^2$ $a = 1-t^2 \Rightarrow \cos x = \frac{1-t^2}{1+t^2}$ * cso | A1 | Correct proof. AO2.1 |

## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $7\sin\theta + 9\cos\theta + 3 = 7\left(\frac{2t}{1+t^2}\right) + 9\left(\frac{1-t^2}{1+t^2}\right) + 3$ | M1 | Uses at least one of $\sin\theta = \frac{2t}{1+t^2}$ or $\cos\theta = \frac{1-t^2}{1+t^2}$ to express in terms of $t$ only. AO1.1b |
| $7\left(\frac{2t}{1+t^2}\right) + 9\left(\frac{1-t^2}{1+t^2}\right) + 3 = 0 \Rightarrow 14t + 9 - 9t^2 + 3 + 3t^2 = 0$ $\Rightarrow 6t^2 - 14t - 12 = 0 \Rightarrow 3t^2 - 7t - 6 = 0 \Rightarrow (t-3)(3t+2) = 0 \Rightarrow t = \ldots$ | M1 | Uses both correct formulae, multiplies by $1+t^2$, forms 3TQ and uses correct method to solve. AO1.1b |
| Either $t = 3 \Rightarrow \frac{\theta}{2} = \arctan(3) \Rightarrow \theta = 2\arctan(3)$ **or** $t = -\frac{2}{3} \Rightarrow \frac{\theta}{2} = 180° + \arctan\left(-\frac{2}{3}\right) \Rightarrow \theta = 2\left(180° + \arctan\left(-\frac{2}{3}\right)\right)$ | M1 | Adopts correct applied strategy to find at least one value of $\theta$ in range $0 < \theta \leq 360°$. AO1.1b |
| $\frac{\theta}{2} = \{71.5650..., 146.3099...\} \Rightarrow \theta = \{143.1301..., 292.6198...\}$ | | |
| $\theta = 143.1°, 292.6°$ (1dp) | A1 | Correct answers only. AO1.1b. Give A0 for extra solutions in range. Ignore extra solutions outside range. |

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