A.Maths(Quadratic equation)

(a)
cos5θ = 0
5θ = 90, 270, 450, 630, 810
θ = 18, 54, 90, 126, 162

(b)
Let z = cosθ + i sinθ
By De Moivre's Theorem, z5 = cos5θ + i sin5θ
By Binomial expension, z5 = cos5θ + 5i cos4θsinθ - 10cos3θsin2θ - 10i cos2θsin3θ + 5cosθsin4θ + isin5θ
Comparing the real part
cos5θ
= cos5θ - 10cos3θsin2θ + 5cosθsin4θ
= cos5θ - 10cos3θ(1 - cos2θ) +5cosθ(1 - cos2θ)2
= 16cos5θ - 20cos3θ + 5cosθ

(ci)
f(x) =16x4 - 20x2 + 5
Putting x=cosθ, f(x) becomes 16cos4θ - 20cos2θ + 5 = cos5θ / cosθ
f(x) = cos5θ / cosθ = 0
i.e. cos5θ = 0 and cosθ does NOT = 0
i.e. θ = 18, 54, 126, 162 for 0°≦θ≦180°
Which follows
f(x)
= 16(x - cos18°)(x - cos162°)(x - cos54°)(x - cos126°)
= 16(x - cos18°)(x + cos18°)(x - cos54°)(x + cos54°)
= 16(x2-cos218°)(x2-cos254°)

(cii)
Putting x2 = y, we have
f(x) = 16(y - cos218°)(y - cos254°) = 16y2 - 20y + 5
This implies the quadratic equation should be g(x) = 16x2 - 20x + 5

Hope the above information helps =)

NOTE: these are questions about complex numbers, which are not included in the present syllabus of additional mathematics.
(a) cos5θ = 0
5θ = 90°, 270°, 450°, 630°, 810°
θ = 18°, 54°, 90°, 126°, 162°
(b) (cosθ + isinθ)^5 = cos5θ + isin5θ by De Moivre's Theorem. Thus
cos5θ + isin5θ
= (cosθ)^5 + 5(cosθ)^4 (isinθ) + 10(cosθ)^3 (isinθ)^2 + 10(cosθ)^2 (isinθ)^3
+ 5(cosθ)(isinθ)^4 + (isinθ)^5
= (cosθ)^5 + 5i(cosθ)^4 (sinθ) - 10(cosθ)^3 (sinθ)^2 - 10i(cosθ)^2 (sinθ)^3
+ 5(cosθ)(sinθ)^4 + i(sinθ)^5
= [(cosθ)^5 - 10(cosθ)^3 (sinθ)^2 + 5(cosθ)(sinθ)^4 ]
+ i[5(cosθ)^4 (sinθ) - 10(cosθ)^2 (sinθ)^3 + (sinθ)^5]
By equating real part, cos5θ = (cosθ)^5 - 10(cosθ)^3 (sinθ)^2 + 5(cosθ)(sinθ)^4
= (cosθ)^5 - 10(cosθ)^3 [1 - (cosθ)^2] + 5(cosθ)[1 - (cosθ)^2]^2
= (cosθ)^5 - 10(cosθ)^3 + 10(cosθ)^5 + 5(cosθ)[1 - 2(cosθ)^2 + (cosθ)^4]
= (cosθ)^5 - 10(cosθ)^3 + 10(cosθ)^5 + 5(cosθ) - 10(cosθ)^3 + 5(cosθ)^5
= 16(cosθ)^5 - 20(cosθ)^3 + 5(cosθ)
(c) (i) Put x = cosθ in f(x) = 0
16(cosθ)^4 - 20(cosθ)^2 + 5 = 0
16(cosθ)^5 - 20(cosθ)^3 + 5(cosθ) = 0(cosθ)
cos5θ = 0 [with cosθ =/= 0]
θ = 18°, 54°, 90°(rejected as cosθ = 0), 126°, 162°
Thus the roots of f(x) = 0 are cos18°, cos54°, cos126°, cos162°
f(x) = A(x - cos18°)(x - cos54°)(x - cos126°)(x - cos162°), where A is a real number.