# Question 9:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\cos 6\theta =)\text{Re}(\cos\theta + i\sin\theta)^6 = \cos^6\theta - 15\cos^4\theta\sin^2\theta + 15\cos^2\theta\sin^4\theta - \sin^6\theta$ | M1A1 | Attempts binomial expansion of $(\cos\theta + i\sin\theta)^6$, extracts real terms; binomial coefficients correct but allow sign errors; allow $(c+is)^6$ notation. A1: correct extracted expression, need not be set equal to $\cos 6\theta$ |
| $= \cos^6\theta - 15\cos^4\theta(1-\cos^2\theta) + 15\cos^2\theta(1-\cos^2\theta)^2 - (1-\cos^2\theta)^3$ | M1 | Applies $\sin^2\theta = 1 - \cos^2\theta$ to obtain expression in $\cos\theta$ only |
| $\cos 6\theta \equiv 32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1$ | A1* | Sets equal to $\cos 6\theta$, fully expands square and cube brackets (must be shown), completes with no errors to obtain given answer |

## Part (a) Alternative:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(z + \frac{1}{z}\right)^6 = z^6 + 6z^4 + 15z^2 + 20 + \frac{15}{z^2} + \frac{6}{z^4} + \frac{1}{z^6}$ | M1A1 | Attempts binomial expansion of $(z+z^{-1})^6$, groups terms, attempts to replace $z^n + z^{-n}$ with cosines; allow errors in "$2\cos(n\theta)$". A1: correct expression in cosines |
| $= 2\cos 6\theta + 6(2\cos 4\theta) + 15(2\cos 2\theta) + 20$ | | |
| $\cos 6\theta \equiv 32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1$ | M1, A1* | M1: applies $\cos 2A = 2\cos^2 A - 1$ repeatedly for $\cos 4\theta$ and $\cos 2\theta$ in terms of $\cos\theta$ only. A1*: sets equal to $64\cos^6\theta$, expands all brackets, rearranges with no errors |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = \cos\theta \Rightarrow \frac{3}{2}\cos 6\theta + \frac{1}{2} = 0 \Rightarrow \cos 6\theta = -\frac{1}{3}$ | M1 | Attempts to use result from part (a) to obtain $\cos 6\theta = k$, $\|k\| < 1$, $k \neq 0$. Note: solving cubic in $x^2$ and square rooting scores no marks |
| $\cos 6\theta = -\frac{1}{3} \Rightarrow 6\theta = \ldots \Rightarrow \theta = \ldots$ (pv $6\theta = 109.5°$, 1.91 rad, $\theta = 18.25°$, 0.318 rad) | dM1 | Solves to obtain at least one value for $\theta$; if $6\theta = \arccos\left(-\frac{1}{3}\right)$ is seen, accept any value that follows; answers must be correct for their $\cos(6\theta)$ if method not shown |
| $x = \cos\theta = \ldots$ (commonly 0.950) | dM1 | Depends on first M; reverses substitution to find at least one value for $x$. Full list: 0.950, 0.746, 0.204, $-0.204$, $-0.746$, $-0.950$ |
| $x = 0.204$ | A1 | Awrt 0.204, must be identified as the answer, not just given in a list |

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