QCE Vault / Specialist Maths Trigonometric proofs using De Moivre — Question 163 Original QCE Vault practice · 6 marks
Browse all questions Original practice exam Report an issue Q163 · Practice question Technology-free Complex familiar 6 marks
QUESTION 163 (6 marks) Consider ( cos θ + i sin θ ) 3 (\cos\theta+i\sin\theta)^3 ( cos θ + i sin θ ) 3 . a) Use De Moivre's theorem to prove that cos 3 θ = 4 cos 3 θ − 3 cos θ \cos3\theta=4\cos^3\theta-3\cos\theta cos 3 θ = 4 cos 3 θ − 3 cos θ . [3 marks] b) Hence solve 4 cos 3 θ − 3 cos θ = 0 4\cos^3\theta-3\cos\theta=0 4 cos 3 θ − 3 cos θ = 0 for 0 ≤ θ < 2 π 0\le\theta<2\pi 0 ≤ θ < 2 π . Give exact values. [3 marks] WORKED SOLUTION
Practice marking scheme 6 marks ANSWER (a) cos 3 θ = 4 cos 3 θ − 3 cos θ \cos3\theta=4\cos^3\theta-3\cos\theta cos 3 θ = 4 cos 3 θ − 3 cos θ . (b) θ = π / 6 , π / 2 , 5 π / 6 , 7 π / 6 , 3 π / 2 , 11 π / 6 \theta=\pi/6,\pi/2,5\pi/6,7\pi/6,3\pi/2,11\pi/6 θ = π /6 , π /2 , 5 π /6 , 7 π /6 , 3 π /2 , 11 π /6 . Worked solution
(a) De Moivre's theorem gives ( cos θ + i sin θ ) 3 = cos 3 θ + i sin 3 θ (\cos\theta+i\sin\theta)^3=\cos3\theta+i\sin3\theta ( cos θ + i sin θ ) 3 = cos 3 θ + i sin 3 θ . Expanding the left side and equating real parts gives cos 3 θ = cos 3 θ − 3 cos θ sin 2 θ = cos 3 θ − 3 cos θ ( 1 − cos 2 θ ) = 4 cos 3 θ − 3 cos θ . \cos3\theta=\cos^3\theta-3\cos\theta\sin^2\theta
=\cos^3\theta-3\cos\theta(1-\cos^2\theta)=4\cos^3\theta-3\cos\theta. cos 3 θ = cos 3 θ − 3 cos θ sin 2 θ = cos 3 θ − 3 cos θ ( 1 − cos 2 θ ) = 4 cos 3 θ − 3 cos θ . (b) The equation is cos 3 θ = 0 \cos3\theta=0 cos 3 θ = 0 . Since 0 ≤ 3 θ < 6 π 0\le3\theta<6\pi 0 ≤ 3 θ < 6 π , 3 θ = π / 2 + k π 3\theta=\pi/2+k\pi 3 θ = π /2 + k π for k = 0 , 1 , 2 , 3 , 4 , 5 k=0,1,2,3,4,5 k = 0 , 1 , 2 , 3 , 4 , 5 . Hence θ = π / 6 + k π / 3 \theta=\pi/6+k\pi/3 θ = π /6 + k π /3 , giving the six values stated. The excluded endpoint 2 π 2\pi 2 π is not included. An equivalent check factors the original expression as cos θ ( 4 cos 2 θ − 3 ) \cos\theta(4\cos^2\theta-3) cos θ ( 4 cos 2 θ − 3 ) . Equivalent mathematically justified methods accepted.
Exact answers unless a decimal accuracy is specified.
Part a: Apply De Moivre and expand the cube.
[1 mark] Part a: Equate real parts.
[1 mark] Part a: Use sin squared plus cos squared equals 1 to prove the identity.
[1 mark] Part b: Rewrite the equation as cos(3 theta)=0.
[1 mark] Part b: Find the six valid arguments for 3 theta.
[1 mark] Part b: List all six theta values in the stated domain.
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