Post a New Question

TRIGONOMETRY

posted by .

Can someone please give me a step by step answer on how to find the complex cube roots of 8(cos(4π/5)+isin(4π/5))? I need to know how to do this for my trig final next week.

  • TRIGONOMETRY -

    We will use De Moivre's theorem

    8(cos(4π/5) + isin(4π/5)^(1/3)
    = 8^(1/3) (cos (1/3)(4π/5) + i sin (1/3)(4π/5) )
    = 2(cos (4π/15) + i sin(4π/15) )

    for a youtube of De Moivre's theorem check out this:

    http://www.youtube.com/watch?v=NrwWv9JHdI4

  • TRIGONOMETRY -

    (rcisθ)^n = r^n cis(nθ)
    cuberoot = 1/3 power, so

    ∛8(cos(4π/5)+isin(4π/5))
    = ∛8 cis((4π/5)/3)
    = 2cis(4π/15)

    Now, since cisθ = cis(θ+2nπ)
    cis 4π/5 = cis 14π/5 = cis 24π/5

    (2cis 4π/15)^3 = 8cis 4π/5
    (2cis (4π/15 + 2π/3))^3 = (2cis 14π/15)^3 = 8cis 14π/5
    (2cis (4π/15 + 4π/3))^3 = (2cis 24π/15)^3 = 8cis 24π/5

    So, you need to evaluate sin & cos of 4π/15,14π/15,24π/15

    adding another 2π takes us around to 34π/15 = 2π + 4π/15 and we're back where we started.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question