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

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.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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

$2(\sin x\cos {60}^{\circ }+\cos x\sin {60}^{\circ })=\cos x\cos {30}^{\circ }-\sin x\sin {30}^{\circ }$     (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 $n=-1,0,1$. Equivalent forms are acceptable.     A1

[5 marks]