The Student Room Group
a) e5iθ=cos5θ+isin5θe^{5i\theta} = \cos 5\theta + i\sin 5\theta
b) e5iθ=(cosθ+isinθ)5e^{5i\theta} = (cos\theta+i\sin\theta)^{5}

Expand out b) and take the imaginary part, which a) tells you will be equal to sin5θ\sin 5\theta. Then divide your expression for sin5θ\sin 5\theta by sin and change any sin's into cos's using sin2θ+cos2θ=1\sin^{2}\theta + \cos^{2}\theta = 1
tribal_angel
How do i use de Moivre''s theorem to prove that, if θ is not a multiple of pi

sin5θ/sinθ = 16 cos^4 θ - 12cos²θ + 1

I could do it if the left hand side was also cos, but i dont know how to go about doing this. Could anyone help?

let c=cosθ s=sinθ
(c+is)^5=c^5+5ic^4s-10c^3s^2-10ic^2s^3+5cs^2+is^5
=cos5θ+isin5θ by De Moivres
comparing imag parts gives
Sin5θ=5cos^4(θ )sinθ-10cos²θsin³θ+sin^5(θ )
dividing by sinθ
=5cos^4(θ )-10cos²θ(1-cos²θ )+(1-cos²θ
=5cos^4(θ ) -10cos²θ+10cos^4(θ )+1-2cos²θ+cos^4(θ )
=16cos^4(θ )-12cos²θ+1