*WILL FAN AND MEDAL* Can someone please take a look at my work and help me work out the rest of this problem? Find the cube roots of 27(cos 330° + i sin 330°).

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*WILL FAN AND MEDAL* Can someone please take a look at my work and help me work out the rest of this problem? Find the cube roots of 27(cos 330° + i sin 330°).

Mathematics
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My work so far:
\[z = r(\cos \theta + i \sin \theta)\]
\[\sqrt[n]{r}(\cos \frac{ \theta+2\pi k }{ n }+i \sin \frac{ \theta+2\pi k }{ n })\]

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Other answers:

I substituted the values into the formula above. K is any real number so I chose to use k= 0, 1, 2. For my first root working with k=0, I got \[3(\cos \frac{ \theta }{ 3 }+i \sin \frac{ \theta }{ 3 })\]
Im stuck on finding the second and third roots..
HI!!
\[\frac{330}{3}=111\] right?
Correct @misty1212
since you seem to be working in degrees, which is weird, but whatever
then to get the next one, go around the circle again
\[330+360=690\] and \[\frac{690}{3}=230\]
so next angle is \(230\) then go around yet again
i can see why it is confusing, because you are using that \(\frac{\theta+2k\pi}{n}\) thing, but you using degrees
if you were using radians, then that is what you would use but since you are using degrees, keep adding 360 then divide not \(2\pi\)
Im using DeMoivre's Theorem, but I was just confused because I'm not sure if I should substitute the angle 330 degrees in place of theta in the expression I have so far. In finding my second root, I used my second value for k, which is 1. Im just stuck on whether or not im supposed to substitute in 330 degrees in place of theta. This is the work I have built around my second root: \[\sqrt[3]{27}(\cos \frac{ \theta+2 \pi(1)}{ 3 }+ i \sin \frac{ \theta+2 \pi(1) }{ 3})\]
I cant give my answers in degrees on this problem though.
Because the standard form of the expression i have is \[z=r(\cos \theta + i \sin \theta)\]
if you are working in degrees, then you have to add 360 not \(2\pi\)
and the problem was given in degrees, so probably you are supposed to answer in degree otherwise it is just too weird
if you want to work in radians, then you would need to convert \(330\) in to radians as step one
For my first root I got \[3(\cos \frac{ \theta }{ 3 } +i \sin \frac{ \theta }{ 3 })\] So if i am to give my answer in degree form, how do I get that from this answer?
Oh ok
\(\theta\) is the angle in this case \(\theta=330\)
Alright, so i substitute 330 in place of theta and simplify?
330/3 = 110
looks like you may be confusing yourself with the formula in simple english to take the cubed root, divide the angle by 3
so first answer is \[3\left(\cos(110^\circ)+i\sin(110^\circ)\right)\]
to get the second answer, as i wrote above, add \(360\) to \(330\) then divide by 3 IF you were working in radians (which you are not) then you would add \(2\pi\) but since you are working in degrees, add \(360\)
I see what you mean here by how confusing it is. So I could solve this just by using a unit circle and finding coterminal angles and dividing them by 3 to get cube roots?
yeah just keep going around
don't forget for the trig form of a complex number the angle is not unique, since sine and cosine are periodic
Okay, so my first root would be 330/3= 110?
no, dear, that is your first ANGLE the root is \[3\left(\cos(110^\circ)+i\sin(110^\circ)\right)\]
OH! Okay, so I do that process but just substitute in my answer for theta?
you have no idea what that number is, since you know nether \(\cos(110^\circ)\) or \(\sin(110^\circ)\) just leave it in that form
ready to find the next root? there are three total
Right, thats what I mean. So my second root would look like this: \[3(\cos(230)+i \sin (230))\]
exactly!
Oh, and then my last root would look like this: 3(cos(196.6)+isin(196.6))
hmmm no
lets back up a second first angle was \(330\div 3=110\)
Yes
second one was \(\frac{330+360}{3}=230\)
Yes
third one add 360 again \[\frac{330+360+360}{3}\]
or more to your formula \[\frac{330+2\times 360}{3}\]
350
you keep going around the circle
right
not sure where the 196.6 came from
Oh my gosh, thank you so much for explaining that. I really needed it. You are by far the most persistent person yet. Thank you so much!
\[\huge \color\magenta\heartsuit\] btw i hope it is now clear who easy this is
Crystal clear, thank you!
you're welcome dear

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