**(a)** $(\cos\theta + i\sin\theta)^4 = \cos 4\theta + i\sin 4\theta$ | B1 |

$(\cos\theta - i\sin\theta)^4 = \cos 4\theta - i\sin 4\theta$ | |

$(c+is)^4 + (c-is)^4 = 2\cos 4\theta$ | M1 |

**Divide throughout by** $\cos^4\theta$ | |

$(1 + i\tan\theta)^4 + (1 - i\tan\theta)^4 = \frac{2\cos 4\theta}{\cos^4\theta}$ | A1 cso | **3 marks** | AG – must see both sides equated; penalise poor notation/brackets for A1cso

**(b)** $\theta = \frac{\pi}{8} \Rightarrow \cos 4\theta = 0$ | | or $\cos 4\theta = 0 \Rightarrow \theta = \frac{\pi}{8}$

$\Rightarrow z = i\tan\frac{\pi}{8}$ is root or satisfies equation | E1 | AG be convinced: must have statement; must mention $i\tan\frac{\pi}{8}$ but may be listed with other 3 roots

$((1+z)^4 + (1-z)^4 = 0)$ | |

other roots are $i\tan\frac{3\pi}{8}, i\tan\frac{5\pi}{8}, i\tan\frac{7\pi}{8}$ | B1 | **2 marks**

**(c)(i)** $\alpha\beta\gamma\delta = i\tan\frac{\pi}{8}\tan\frac{3\pi}{8}\tan\frac{5\pi}{8}\tan\frac{7\pi}{8}$ | M1 | product of their 4 roots

$\tan\frac{5\pi}{8} = -\tan\frac{3\pi}{8}$ and $\tan\frac{7\pi}{8} = -\tan\frac{\pi}{8}$ | B1 | May earn this mark in part (c)(ii) if not earned here

$((1+z)^4 + (1-z)^4 = 0)$ | | or $z^4 + 6z^2 + 1 = (0)$ seen

$\alpha\beta\gamma\delta = 1 \Rightarrow \tan\frac{\pi}{8}\tan\frac{3\pi}{8} = 1$ | A1 cso | **4 marks** | must see $i^4$ become 1 for final A1 cso

**(c)(ii)** $\left(\sum\alpha\right)^2 = \sum\alpha^2 + 2\sum\alpha\beta$ | M1 |

$\sum\alpha = 0 = \sum\alpha^2 = -2\sum\alpha\beta = -12$ | A1 | using $z^4 + 6z^2 + 1 = 0$

$i^2\left(\tan\frac{\pi}{8} + \tan\frac{3\pi}{8} + \tan\frac{5\pi}{8} + \tan\frac{7\pi}{8}\right) = -12$ OE | A1 | $\tan^2\frac{\pi}{8} + \tan^2\frac{3\pi}{8} + \tan^2\frac{5\pi}{8} + \tan^2\frac{7\pi}{8} = 12$ OE

$\tan^2\frac{\pi}{8} + \tan^2\frac{3\pi}{8} = 6$ | A1 cso | **4 marks** | must see $i^2$ become $-1$ for final A1 cso

**Total: 13 marks**

**Notes:**
- **(a)** May also earn M1 for both $(1 + i\tan\theta)^4 = \frac{(\cos\theta + i\sin\theta)^4}{\cos^4\theta}$ or $\frac{\cos 4\theta + i\sin 4\theta}{\cos^4\theta}$ and A1 for completing the proof
- Provided de Moivre's theorem is used, award M1 for showing either $\frac{2\cos 4\theta}{\cos^4\theta} = 2 - 12\tan^2\theta + 2\tan^4\theta$ or $(1 + i\tan\theta)^4 + (1 - i\tan\theta)^4 = 2 - 12\tan^2\theta + 2\tan^4\theta$ and A1 for completing the proof
- **(c)(i)** Must use equations in $z$ and roots of form $\tan\alpha$ to earn marks in part (c). Condone omission of all 4 $i$'s for M1 but withhold **A1cso** unless $i^4 = 1$ is seen

**see next page for alternative solution when candidates answer part (c) holistically by converting the quartic equation into a quadratic equation**

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## Question 8 - Alternative Solution:

**(c)** **Alternative part (c)**

**Substitute** $y = z^2$ | M1 |

$(1+z)^4 + (1-z)^4 = 0$ **becomes** | |

$(2)(y^2 + 6y + 1) = 0$ | A1 |

$\tan\frac{5\pi}{8} = -\tan\frac{3\pi}{8}$ and $\tan\frac{7\pi}{8} = -\tan\frac{\pi}{8}$ | B1 |

**Roots are** $-\tan^2\frac{\pi}{8}$ and $-\tan^2\frac{3\pi}{8}$ | E1 | explicitly stated and evidence that $i^2 = -1$ has been used

**Sum of roots is** $-6$ | m1 | FT their quadratic

$\tan^2\frac{\pi}{8} + \tan^2\frac{3\pi}{8} = 6$ | A1 cso | must have earned E1

| | **8 marks** |

**Product of roots is** $1$ | m1 |

$\tan^2\frac{\pi}{8}\tan^2\frac{3\pi}{8} = 1$ | A1 cso | must have earned E1

| | **8 marks** |

**Mark holistically out of 8 and then allocate marks by giving up to 4 marks in (c)(i) and the remainder in part (c)(ii)**

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