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]