## (i)(a)

$\sin\frac{\pi}{8} = \frac{1}{\sqrt{2}}, \sin\frac{5\pi}{8} = \frac{1}{\sqrt{2}}$

| B1 | For verifying $\theta = \frac{1}{8}\pi$ (1 mark) |

## (i)(b)

[Graph showing $y = \sin 6\theta$ and $y = \sin 2\theta$ for $0, 0, \frac{1}{2}\pi$]

| M1 | For sketching $y = \sin 6\theta$ and $y = \sin 2\theta$ for $0, 0, \frac{1}{2}\pi$ OR any other correct method for solving $\sin 6\theta = \sin 2\theta$ for $\theta \neq k\frac{\pi}{3}$ OR appropriate use of symmetry OR attempt to verify a reasonable guess for $\theta$ |
| A1 | For correct $\theta$ (2 marks) |

$\theta = \frac{3}{8}\pi$

## (ii)

$\text{Im}(c + is)^6 = 6c^5s - 20c^3s^3 + 6cs^5$

| M1 | For expanding $(c+is)^6$; at least 3 terms and 3 binomial coefficients needed |
| A1 | For 3 correct terms |

$\sin 6\theta = \sin\theta(6c^5 - 20c^3(1-c^2) + 6c(1-c^2)^2)$

| M1 | For using $s^2 = 1 - c^2$ |
| A1 | For any correct intermediate stage |

$\sin 6\theta = \sin\theta(32c^5 - 32c^3 + 6c)$

| A1 | For obtaining this expression correctly |

$\sin 6\theta = 2\sin\theta\cos\theta(16c^4 - 16c^2 + 3)$

$\sin 6\theta = \sin 2\theta[16\cos^4\theta - 16\cos^2\theta + 3]$

| AG | (5 marks) |

## (iii)

$16c^4 - 16c^2 + 3 = 1$

$\Rightarrow c^2 = \frac{2 \pm \sqrt{2}}{4}$

| M1 | For stating this equation AEF |
| A1 | For obtaining both values of $c^2$ |

$-$ sign requires larger $\theta = \frac{3}{8}\pi$

| A1 | For stating and justifying $\theta = \frac{3}{8}\pi$ Calculator OK if figures seen (3 marks) |

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