Mathematical induction and trigonometric proofs — Question 178
Original QCE Vault practice · 6 marks
Q178 · Practice questionTechnology-freeComplex familiar6 marks
QUESTION 178 (6 marks)
Consider (cosθ+isinθ)4. a)Use De Moivre's theorem to prove cos4θ=8cos4θ−8cos2θ+1. [3 marks] b)Hence solve 8cos4θ−8cos2θ+1=−1/2 for 0≤θ≤π/2. [3 marks] WORKED SOLUTION
Practice marking scheme
6 marksANSWER(a) Identity proved. (b) θ=π/6,π/3. Worked solution
(a) De Moivre gives real part cos4θ. Expansion gives cos4θ−6cos2θsin2θ+sin4θ. Substitute sin2θ=1−cos2θ and expand to obtain 8cos4θ−8cos2θ+1. (b) The equation is cos4θ=−1/2. Since 0≤4θ≤2π, 4θ=2π/3 or 4π/3. Division by 4 gives the two stated values; neither endpoint solves the equation. Equivalent mathematically justified methods accepted.
Exact answers unless a decimal accuracy is specified.
Part a: Apply De Moivre and expand.
[1 mark]Part a: Equate real parts.
[1 mark]Part a: Simplify using the Pythagorean identity.
[1 mark]Part b: Rewrite as cos(4 theta)=-1/2.
[1 mark]Part b: Find both valid transformed arguments.
[1 mark]Part b: Give both theta values.
[1 mark]Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
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