# Question 8:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $z^n = e^{in\theta} = \cos n\theta + i\sin n\theta$ | | |
| $\frac{1}{z^n} = e^{-in\theta} = \cos n\theta - i\sin n\theta$ | | |
| $z^n + \frac{1}{z^n} = 2\cos n\theta$ | M1A1cso | (2) Attempt to obtain $z^n + \frac{1}{z^n}$; reach given result with clear working; must see $\cos(-n\theta)+i\sin(-n\theta)$ changed to $\cos n\theta - i\sin n\theta$ |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(z+\frac{1}{z}\right)^6 = z^6 + 6z^5\cdot\frac{1}{z} + \frac{6\times5}{2!}z^4\cdot\frac{1}{z^2} + \frac{6\times5\times4}{3!}z^3\cdot\frac{1}{z^3} + \frac{6\times5\times4\times3}{4!}z^2\cdot\frac{1}{z^4} + \frac{6\times5\times4\times3\times2}{5!}z\cdot\frac{1}{z^5} + \frac{1}{z^6}$ | M1A1 | Apply binomial expansion; coefficients must be numerical; expansion must have 7 terms with at least 4 correct; terms need not be simplified |
| $(2\cos\theta)^6 = z^6 + 6z^4 + 15z^2 + 20 + 15\cdot\frac{1}{z^2} + 6\cdot\frac{1}{z^4} + \frac{1}{z^6}$ | M1 | Simplify coefficients and pair appropriate terms; at least 2 pairs correct |
| $64\cos^6\theta = z^6 + \frac{1}{z^6} + 6\!\left(z^4+\frac{1}{z^4}\right) + 15\!\left(z^2+\frac{1}{z^2}\right) + 20$ | M1 | Use result from (a); must include $2^6$ or 64 |
| $64\cos^6\theta = 2\cos 6\theta + 6\times 2\cos 4\theta + 15\times 2\cos 2\theta + 20$ | M1 | |
| $\cos^6\theta = \frac{1}{32}(\cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10)$ | A1* | (5) Obtain given result with no errors in working |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10 = 10$ | M1A1 | Use result from (b) to simplify given equation; reach $32\cos^6\theta = 10$ |
| $\cos\theta = \pm\sqrt[6]{\frac{5}{16}}$ | M1 | Solve to obtain at least one correct value for $\theta$ in radians in given range, 3 sf or better |
| $\theta = 0.603,\ 2.54$ | A1 | (4) 2 correct values and no extras in radians in range; must be 3 sf; ignore extras outside range |

## Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_0^{\frac{\pi}{3}}(32\cos^6\theta - 4\cos^2\theta)\,d\theta$ | | |
| $= \int_0^{\frac{\pi}{3}}(\cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10 - 4\cos^2\theta)\,d\theta$ | M1 | Use result in (b) to change $\cos^6\theta$ to sum of multiple angles; use $\cos^2\theta = \pm\frac{1}{2}(\cos 2\theta \pm 1)$; limits not needed |
| $= \int_0^{\frac{\pi}{3}}(\cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10 - 2 - 2\cos 2\theta)\,d\theta$ | | |
| $= \left[\frac{1}{6}\sin 6\theta + \frac{3}{2}\sin 4\theta + \frac{13}{2}\sin 2\theta + 8\theta\right]_0^{\frac{\pi}{3}}$ | M1A1 | Integrate to obtain terms in $\sin 6\theta, \sin 4\theta, \sin 2\theta$ and $\theta$; correct integration |
| $= (0) + \frac{3}{2}\!\left(-\frac{\sqrt{3}}{2}\right) + \frac{13}{2}\times\frac{\sqrt{3}}{2} + \frac{8\pi}{3} - (-0)$ | dM1 | Substitute limit $\frac{\pi}{3}$; depends on second M mark |
| $= \frac{5\sqrt{3}}{2} + \frac{8\pi}{3}$ | A1 | (5) Correct exact answer or equivalent **[16]** |

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