Question: show that Sin(5θ) = 5sin(θ) -20sin^3(θ) + 16sin^5(θ).
By De Moivre's
(cosθ + i.sinθ)^5 = cos(5θ) + i.sin(5θ)
By expansion
(cosθ + i.sinθ)^5 = cos^5(θ) + 5.cos^4(θ).i.sin(θ) + 10.cos^3(θ).i^2.sin^2(θ) + 10.cos^2(θ).i^3.sin^3(θ) + 5.cos(θ).i^4.sin^4(θ) + i^5.sin^5(θ)
Comparing the imaginary parts of (rhs) both expressions,
i.sin(5θ) = i(5.cos^4(θ).sin(θ) - 10.cos^2(θ).sin^3(θ) + sin^5(θ))
sin(5θ) = 5.(1 - sin^2(θ))^2.sin(θ) - 10.(1 - sin^2(θ)).sin^3(θ) + sin^5(θ)
sin(5θ) = 5.(1 - 2.sin^2(θ) + sin^4(θ)).sin(θ) - 10.(sin^3(θ) - sin^5(θ)) + sin^5(θ)
sin(5θ) = 5.(sin(θ) - 2.sin^3(θ) + sin^5(θ)) - 10.(sin^3(θ) - sin^5(θ)) + sin^5(θ)
sin(5θ) = 5.sin(θ) - 10.sin^3(θ) + 5.sin^5(θ) - 10.sin^3(θ) + 10.sin^5(θ) + sin^5(θ)
sin(5θ) = 5.sin(θ) - 20.sin^3(θ) + 16.sin^5(θ)