Solve $2\sin (x + 60^\circ ) = \cos (x + 30^\circ ),{\text{ }}0^\circ \leqslant x \leqslant 180^\circ$.

Show that $\sin 105^\circ + \cos 105^\circ = \frac{1}{{\sqrt 2 }}$.

Find the modulus and argument of $z$ in terms of $\theta$. Express each answer in its simplest form.

Hence find the cube roots of $z$ in modulus-argument form.

$2\sin (x + 60^\circ ) = \cos (x + 30^\circ )$

$2(\sin x\cos 60^\circ + \cos x\sin 60^\circ ) = \cos x\cos 30^\circ - \sin x\sin 30^\circ$     (M1)(A1)

$2\sin x \times \frac{1}{2} + 2\cos x \times \frac{{\sqrt 3 }}{2} = \cos x \times \frac{{\sqrt 3 }}{2} - \sin x \times \frac{1}{2}$     A1

$\Rightarrow \frac{3}{2}\sin x = - \frac{{\sqrt 3 }}{2}\cos x$

$\Rightarrow \tan x = - \frac{1}{{\sqrt 3 }}$     M1

$\Rightarrow x = 150^\circ$     A1

[5 marks]

EITHER

choosing two appropriate angles, for example 60° and 45°     M1

$\sin 105^\circ = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ$ and

$\cos 105^\circ = \cos 60^\circ \cos 45^\circ - \sin 60^\circ \sin 45^\circ$     (A1)

$\sin 105^\circ + \cos 105^\circ = \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 2 }} + \frac{1}{2} \times \frac{1}{{\sqrt 2 }} + \frac{1}{2} \times \frac{1}{{\sqrt 2 }} - \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 2 }}$     A1

$= \frac{1}{{\sqrt 2 }}$     AG

OR

attempt to square the expression     M1

${(\sin 105^\circ + \cos 105^\circ )^2} = {\sin ^2}105^\circ + 2\sin 105^\circ \cos 105^\circ + {\cos ^2}105^\circ$

${(\sin 105^\circ + \cos 105^\circ )^2} = 1 + \sin 210^\circ$     A1

$= \frac{1}{2}$     A1

$\sin 105^\circ + \cos 105^\circ = \frac{1}{{\sqrt 2 }}$   AG

[3 marks]

EITHER

$z = (1 - \cos 2\theta ) - {\text{i}}\sin 2\theta$

$\left| z \right| = \sqrt {{{(1 - \cos 2\theta )}^2} + {{(\sin 2\theta )}^2}}$     M1

$\left| z \right| = \sqrt {1 - 2\cos 2\theta + {{\cos }^2}2\theta + {{\sin }^2}2\theta }$     A1

$= \sqrt 2 \sqrt {(1 - \cos 2\theta )}$     A1

$= \sqrt {2(2{{\sin }^2}\theta )}$

$= 2\sin \theta$     A1

let $\arg (z) = \alpha$

$\tan \alpha = - \frac{{\sin 2\theta }}{{1 - \cos 2\theta }}$     M1

$= \frac{{ - 2\sin \theta \cos \theta }}{{2{{\sin }^2}\theta }}$     (A1)

$= - \cot \theta$     A1

$\arg (z) = \alpha = - \arctan \left( {\tan \left( {\frac{\pi }{2} - \theta } \right)} \right)$     A1

$= \theta - \frac{\pi }{2}$     A1

OR

$z = (1 - \cos 2\theta ) - {\text{i}}\sin 2\theta$

$= 2{\sin ^2}\theta - 2{\text{i}}\sin \theta \cos \theta$     M1A1

$= 2\sin \theta (\sin \theta - {\text{i}}\cos \theta )$     (A1)

$= - 2{\text{i}}\sin \theta (\cos \theta + {\text{i}}\sin \theta )$     M1A1

$= 2\sin \theta \left( {\cos \left( {\theta - \frac{\pi }{2}} \right) + {\text{i}}\sin \left( {\theta - \frac{\pi }{2}} \right)} \right)$     M1A1

$\left| z \right| = 2\sin \theta$     A1

$\arg (z) = \theta - \frac{\pi }{2}$     A1

[9 marks]

attempt to apply De Moivre’s theorem     M1

${(1 - \cos 2\theta - {\text{i}}\sin 2\theta )^{\frac{1}{3}}} = {2^{\frac{1}{3}}}{(\sin \theta )^{\frac{1}{3}}}\left[ {\cos \left( {\frac{{\theta - \frac{\pi }{2} + 2n\pi }}{3}} \right) + {\text{i}}\sin \left( {\frac{{\theta - \frac{\pi }{2} + 2n\pi }}{3}} \right)} \right]$     A1A1A1

Note:     A1 for modulus, A1 for dividing argument of $z$ by 3 and A1 for $2n\pi$.

Hence cube roots are the above expression when $n = - 1,{\text{ }}0,{\text{ }}1$. Equivalent forms are acceptable.     A1

[5 marks]