Let $w = \cos \frac{{2\pi }}{5} + {\text{i}}\sin \frac{{2\pi }}{5}$.

(a)     Show that w is a root of the equation ${z^5} - 1 = 0$ .

(b)     Show that $(w - 1)({w^4} + {w^3} + {w^2} + w + 1) = {w^5} - 1$ and deduce that ${w^4} + {w^3} + {w^2} + w + 1 = 0$.

(c)     Hence show that $\cos \frac{{2\pi }}{5} + \cos \frac{{4\pi }}{5} = - \frac{1}{2}$.

(a)     EITHER

${w^5} = {\left( {\cos \frac{{2\pi }}{5} + {\text{i}}\sin \frac{{2\pi }}{5}} \right)^5}$     (M1)

$= \cos 2\pi + {\text{i}}\sin 2\pi$     A1

$= 1$     A1

Hence w is a root of ${z^5} - 1 = 0$     AG

OR

Solving ${z^5} = 1$     (M1)

$z = \cos \frac{{2\pi }}{5}n + {\text{i}}\sin \frac{{2\pi }}{5}n{\text{ ,}}\,\,\,\,\,n = {\text{0, 1, 2, 3, 4}}$.     A1

$n = 1{\text{ gives }}\cos \frac{{2\pi }}{5} + {\text{i}}\sin \frac{{2\pi }}{5}$ which is w     A1

[3 marks]

 

(b)     $(w - 1)(1 + w + {w^2} + {w^3} + {w^4}) = w + {w^2} + {w^3} + {w^4} + {w^5} - 1 - w - {w^2} - {w^3} - {w^4}$     M1

$= {w^5} - 1$     A1

Since ${w^5} - 1 = 0$ and $w \ne 1{\text{ , }}{w^4} + {w^3} + {w^2} + w + 1 = 0$.     R1

[3 marks]

 

(c)     $1 + w + {w^2} + {w^3} + {w^4} =$

$1 + \cos \frac{{2\pi }}{5} + {\text{i}}\sin \frac{{2\pi }}{5} + {\left( {\cos \frac{{2\pi }}{5} + {\text{i}}\sin \frac{{2\pi }}{5}} \right)^2} + {\left( {\cos \frac{{2\pi }}{5} + {\text{i}}\sin \frac{{2\pi }}{5}} \right)^3} + {\left( {\cos \frac{{2\pi }}{5} + {\text{i}}\sin \frac{{2\pi }}{5}} \right)^4}$     (M1)

$= 1 + \cos \frac{{2\pi }}{5} + {\text{i}}\sin \frac{{2\pi }}{5} + \cos \frac{{4\pi }}{5} + {\text{i}}\sin \frac{{4\pi }}{5} + \cos \frac{{6\pi }}{5} + {\text{i}}\sin \frac{{6\pi }}{5} + \cos \frac{{8\pi }}{5} + {\text{i}}\sin \frac{{8\pi }}{5}$     M1

$= 1 + \cos \frac{{2\pi }}{5} + {\text{i}}\sin \frac{{2\pi }}{5} + \cos \frac{{4\pi }}{5} + {\text{i}}\sin \frac{{4\pi }}{5} + \cos \frac{{4\pi }}{5} - {\text{i}}\sin \frac{{4\pi }}{5} + \cos \frac{{2\pi }}{5} - {\text{i}}\sin \frac{{2\pi }}{5}$     M1A1A1

Note: Award M1 for attempting to replace ${6\pi }$ and ${8\pi }$ by ${4\pi }$ and ${2\pi }$ .

Award A1 for correct cosine terms and A1 for correct sine terms.

 

$= 1 + 2\cos \frac{{4\pi }}{5} + 2\cos \frac{{2\pi }}{5} = 0$     A1

Note: Correct methods involving equating real parts, use of conjugates or reciprocals are also accepted.

 

$\cos \frac{{2\pi }}{5} + \cos \frac{{4\pi }}{5} = - \frac{1}{2}$     AG

[6 marks]

Note: Use of cis notation is acceptable throughout this question.

 

Total [12 marks]

Parts (a) and (b) were generally well done, although very few stated that $w \ne 1$ in (b). Part (c), the last question on the paper was challenging. Those candidates who gained some credit correctly focussed on the real part of the identity and realise that different cosine were related.