(a) De Moivre’s theorem gives (cosx+isinx)4=cos4x+isin4x. Expanding and equating real parts yields cos4x=cos4x−6cos2xsin2x+sin4x. Substitute sin2x=1−cos2x: cos4x=cos4x−6cos2x(1−cos2x)+(1−cos2x)2=8cos4x−8cos2x+1.
(b) On the stated interval, 0≤4x≤2π. Thus 4x=π/2 or 3π/2, giving x=π/8 or 3π/8.
Expands the fourth power.
[2 marks]
Equates real parts and simplifies.
[2 marks]
Identifies the two values of 4x.
[1 mark]
Obtains both values of x.
[1 mark]
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.