Question

Find the cube roots of 27(cos 279° + i sin 279°).

Answer 1

@Hero

Answer 2

@Vocaloid

Answer 3

r^(1/n) * [cos(0 + 2pi *k)/n + i * sin(0 + 2pi*k)/n)

where n = 3 (since we have a cube root), r = 27, and k = 0, 1, and 2

you must plug in k = 0, 1, and 2 separately to generate three roots

Answer 4

im supposed to show how i get the answer though like step by step can you help me with that

Answer 5

like i know the answer i just dont know how to explain it all

Answer 6

the answer i get is 3(cos 93 + i sin 93)

Answer 7

well first we start w/ the formula

r^(1/n) * [cos(0 + 2pi *k)/n + i * sin(0 + 2pi*k)/n)

i don't think they'd want you to derive this formula, it's pretty complex

plugging in r and n gives us

27^(1/3) * [cos(0 + 2pi *k)/3 + i * sin(0 + 2pi*k)/3)

then first root: k = 0 gives us

27^(1/3) * [cos(0 + 2pi *0)/3 + i * sin(0 + 2pi*0)/3)

second root k = 1 gives us

27^(1/3) * [cos(0 + 2pi *1)/3 + i * sin(0 + 2pi*1)/3)

third root k = 2 gives us

27^(1/3) * [cos(0 + 2pi *2)/3 + i * sin(0 + 2pi*2)/3)

showing you plugging in the numbers and doing any algebraic simplification should be sufficient

Answer 8

So what do i put just that formula with me explaining what i plugged in?

Answer 9

yes

Answer 10

Thank you!

Answer 11

So our formula we use is,
r^(1/n) * [cos(0 + 2pi *k)/n + i * sin(0 + 2pi*k)/n)

So first we plug in r and n gives which gives us this,
27^(1/3) * [cos(0 + 2pi *k)/3 + i * sin(0 + 2pi*k)/3)

So now we plug in all of our roots (0,1,2)
k = 0 gives us
27^(1/3) * [cos(0 + 2pi *0)/3 + i * sin(0 + 2pi*0)/3)

k = 1 gives us
27^(1/3) * [cos(0 + 2pi *1)/3 + i * sin(0 + 2pi*1)/3)

k = 2 gives us
27^(1/3) * [cos(0 + 2pi *2)/3 + i * sin(0 + 2pi*2)/3)

So our cube roots for this equation are (0,1, and 2)

Answer 12

does that look good?

Answer 13

yes

they probably expect you to simplify the angles, too

ex: cos(0 + 2pi *0)/3 becomes cos (0)

Answer 14

hold on

Answer 15

ok

Answer 16

ah, I forgot the theta value they give you is 279 not 0

then first root: k = 0 gives us

27^(1/3) * [cos(279 + 2pi *0)/3 + i * sin(0 + 2pi*0)/3)

second root k = 1 gives us

27^(1/3) * [cos(279 + 2pi *1)/3 + i * sin(0 + 2pi*1)/3)

third root k = 2 gives us

27^(1/3) * [cos(279 + 2pi *2)/3 + i * sin(0 + 2pi*2)/3)

Answer 17

ok sweet thanks