Write down the expansion of ${\left( {\cos \theta + {\text{i}}\sin \theta } \right)^3}$ in the form $a + {\text{i}}b$ , where $a$ and $b$ are in terms of ${\sin \theta }$ and ${\cos \theta }$ .

Hence show that $\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta$ .

Similarly show that $\cos 5\theta = 16{\cos ^5}\theta - 20{\cos ^3}\theta + 5\cos \theta$ .

Hence solve the equation $\cos 5\theta + \cos 3\theta + \cos \theta = 0$ , where $\theta \in \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$ .

By considering the solutions of the equation $\cos 5\theta = 0$ , show that $\cos \frac{\pi }{{10}} = \sqrt {\frac{{5 + \sqrt 5 }}{8}}$ and state the value of $\cos \frac{{7\pi }}{{10}}$.

${\left( {\cos \theta + {\text{i}}\sin \theta } \right)^3} = {\cos ^3}\theta + 3{\cos ^2}\theta \left( {{\text{i}}\sin \theta } \right) + 3\cos \theta {\left( {{\text{i}}\sin \theta } \right)^2} + {\left( {{\text{i}}\sin \theta } \right)^3}$     (M1)

$= {\cos ^3}\theta - 3\cos \theta {\sin ^2}\theta + {\text{i}}\left( {3{{\cos }^2}\theta \sin \theta - {{\sin }^3}\theta } \right)$     A1

[2 marks]

from De Moivre’s theorem

${\left( {\cos \theta + {\text{i}}\sin \theta } \right)^3} = \cos 3\theta + {\text{i}}\sin 3\theta$     (M1)

$\cos 3\theta + {\text{i}}\sin 3\theta = \left( {{{\cos }^3}\theta - 3\cos \theta {{\sin }^2}\theta } \right) + {\text{i}}\left( {3{{\cos }^2}\theta \sin \theta - {{\sin }^3}\theta } \right)$

equating real parts     M1

$\cos 3\theta = {\cos ^3}\theta - 3\cos \theta {\sin ^2}\theta$

$= {\cos ^3}\theta - 3\cos \theta \left( {1 - {{\cos }^2}\theta } \right)$     A1

$= {\cos ^3}\theta - 3\cos \theta + 3{\cos ^3}\theta$

$= 4{\cos ^3}\theta - 3\cos \theta$     AG

Note: Do not award marks if part (a) is not used.

[3 marks]

${\left( {\cos \theta + {\text{i}}\sin \theta } \right)^5} =$

${\cos ^5}\theta + 5{\cos ^4}\theta \left( {{\text{i}}\sin \theta } \right) + 10{\cos ^3}\theta {\left( {{\text{i}}\sin \theta } \right)^2} + 10{\cos ^2}\theta {\left( {{\text{i}}\sin \theta } \right)^3} + 5\cos \theta {\left( {{\text{i}}\sin \theta } \right)^4} + {\left( {{\text{i}}\sin \theta } \right)^5}$     (A1)

from De Moivre’s theorem

$\cos 5\theta = {\cos ^5}\theta - 10{\cos ^3}\theta {\sin ^2}\theta  + 5\cos \theta {\sin ^4}\theta$     M1

$= {\cos ^5}\theta - 10{\cos ^3}\theta \left( {1 - {{\cos }^2}\theta } \right) + 5\cos \theta {\left( {1 - {{\cos }^2}\theta } \right)^2}$     A1

$= {\cos ^5}\theta - 10{\cos ^3}\theta + 10{\cos ^5}\theta + 5\cos \theta - 10{\cos ^3}\theta + 5{\cos ^5}\theta$

$\therefore \cos 5\theta = 16{\cos ^5}\theta - 20{\cos ^3}\theta + 5\cos \theta$     AG

Note: If compound angles used in (b) and (c), then marks can be allocated in (c) only.

[3 marks]

$\cos 5\theta + \cos 3\theta + \cos \theta$

$= \left( {16{{\cos }^5}\theta - 20{{\cos }^3}\theta + 5\cos \theta } \right) + \left( {4{{\cos }^3}\theta - 3\cos \theta } \right) + \cos \theta = 0$     M1

$16{\cos ^5}\theta - 16{\cos ^3}\theta + 3\cos \theta = 0$     A1

$\cos \theta \left( {16{{\cos }^4}\theta - 16{{\cos }^2}\theta + 3} \right) = 0$

$\cos \theta \left( {4{{\cos }^2}\theta - 3} \right)\left( {4{{\cos }^2}\theta - 1} \right) = 0$     A1

$\therefore \cos \theta = 0$; $\pm \frac{{\sqrt 3 }}{2}$; $\pm \frac{1}{2}$     A1

$\therefore \theta = \pm \frac{\pi }{6}$; $\pm \frac{\pi }{3}$; $\pm \frac{\pi }{2}$     A2

[6 marks]

$\cos 5\theta = 0$

$5\theta = ...\frac{\pi }{2}$; $\left( {\frac{{3\pi }}{2};\frac{{5\pi }}{2}} \right)$; $\frac{{7\pi }}{2}$; $...$     (M1)

$\theta = ...\frac{\pi }{{10}}$; $\left( {\frac{{3\pi }}{{10}};\frac{{5\pi }}{{10}}} \right)$; $\frac{{7\pi }}{10}$; $...$     (M1)

Note: These marks can be awarded for verifications later in the question.

now consider $16{\cos ^5}\theta - 20{\cos ^3}\theta + 5\cos \theta = 0$     M1

$\cos \theta \left( {16{{\cos }^4}\theta - 20{{\cos }^2}\theta + 5} \right) = 0$

${\cos ^2}\theta = \frac{{20 \pm \sqrt {400 - 4\left( {16} \right)\left( 5 \right)} }}{{32}}$; $\cos \theta = 0$     A1

$\cos \theta = \pm \sqrt {\frac{{20 \pm \sqrt {400 - 4\left( {16} \right)\left( 5 \right)} }}{{32}}}$

$\cos \frac{\pi }{{10}} = \sqrt {\frac{{20 + \sqrt {400 - 4\left( {16} \right)\left( 5 \right)} }}{{32}}}$ since max value of cosine
$\Rightarrow$ angle closest to zero     R1

$\cos \frac{\pi }{{10}} = \sqrt {\frac{{4.5 + 4\sqrt {25 - 4\left( 5 \right)} }}{{4.8}}}  = \sqrt {\frac{{5 + \sqrt 5 }}{8}}$     A1

$\cos \frac{{7\pi }}{{10}} = - \sqrt {\frac{{5 - \sqrt 5 }}{8}}$     A1A1

[8 marks]

This question proved to be very difficult for most candidates. Many had difficulties in following the instructions and attempted to use addition formulae rather than binomial expansions. A small number of candidates used the results given and made a good attempt to part (d) but very few answered part (e).

This question proved to be very difficult for most candidates. Many had difficulties in following the instructions and attempted to use addition formulae rather than binomial expansions. A small number of candidates used the results given and made a good attempt to part (d) but very few answered part (e).

This question proved to be very difficult for most candidates. Many had difficulties in following the instructions and attempted to use addition formulae rather than binomial expansions. A small number of candidates used the results given and made a good attempt to part (d) but very few answered part (e).

This question proved to be very difficult for most candidates. Many had difficulties in following the instructions and attempted to use addition formulae rather than binomial expansions. A small number of candidates used the results given and made a good attempt to part (d) but very few answered part (e).

This question proved to be very difficult for most candidates. Many had difficulties in following the instructions and attempted to use addition formulae rather than binomial expansions. A small number of candidates used the results given and made a good attempt to part (d) but very few answered part (e).