anonymous
  • anonymous
*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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
My work so far:
anonymous
  • anonymous
\[z = r(\cos \theta + i \sin \theta)\]
anonymous
  • anonymous
\[\sqrt[n]{r}(\cos \frac{ \theta+2\pi k }{ n }+i \sin \frac{ \theta+2\pi k }{ n })\]

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anonymous
  • anonymous
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 \[3(\cos \frac{ \theta }{ 3 }+i \sin \frac{ \theta }{ 3 })\]
anonymous
  • anonymous
Im stuck on finding the second and third roots..
misty1212
  • misty1212
HI!!
misty1212
  • misty1212
\[\frac{330}{3}=111\] right?
anonymous
  • anonymous
Correct @misty1212
misty1212
  • misty1212
since you seem to be working in degrees, which is weird, but whatever
misty1212
  • misty1212
then to get the next one, go around the circle again
misty1212
  • misty1212
\[330+360=690\] and \[\frac{690}{3}=230\]
misty1212
  • misty1212
so next angle is \(230\) then go around yet again
misty1212
  • misty1212
i can see why it is confusing, because you are using that \(\frac{\theta+2k\pi}{n}\) thing, but you using degrees
misty1212
  • misty1212
if you were using radians, then that is what you would use but since you are using degrees, keep adding 360 then divide not \(2\pi\)
anonymous
  • anonymous
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: \[\sqrt[3]{27}(\cos \frac{ \theta+2 \pi(1)}{ 3 }+ i \sin \frac{ \theta+2 \pi(1) }{ 3})\]
anonymous
  • anonymous
I cant give my answers in degrees on this problem though.
anonymous
  • anonymous
Because the standard form of the expression i have is \[z=r(\cos \theta + i \sin \theta)\]
misty1212
  • misty1212
if you are working in degrees, then you have to add 360 not \(2\pi\)
misty1212
  • misty1212
and the problem was given in degrees, so probably you are supposed to answer in degree otherwise it is just too weird
misty1212
  • misty1212
if you want to work in radians, then you would need to convert \(330\) in to radians as step one
anonymous
  • anonymous
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?
anonymous
  • anonymous
Oh ok
misty1212
  • misty1212
\(\theta\) is the angle in this case \(\theta=330\)
anonymous
  • anonymous
Alright, so i substitute 330 in place of theta and simplify?
anonymous
  • anonymous
330/3 = 110
misty1212
  • misty1212
looks like you may be confusing yourself with the formula in simple english to take the cubed root, divide the angle by 3
misty1212
  • misty1212
so first answer is \[3\left(\cos(110^\circ)+i\sin(110^\circ)\right)\]
misty1212
  • misty1212
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 \(360\)
anonymous
  • anonymous
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?
misty1212
  • misty1212
yeah just keep going around
misty1212
  • misty1212
don't forget for the trig form of a complex number the angle is not unique, since sine and cosine are periodic
anonymous
  • anonymous
Okay, so my first root would be 330/3= 110?
misty1212
  • misty1212
no, dear, that is your first ANGLE the root is \[3\left(\cos(110^\circ)+i\sin(110^\circ)\right)\]
anonymous
  • anonymous
OH! Okay, so I do that process but just substitute in my answer for theta?
misty1212
  • misty1212
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
misty1212
  • misty1212
ready to find the next root? there are three total
anonymous
  • anonymous
Right, thats what I mean. So my second root would look like this: \[3(\cos(230)+i \sin (230))\]
misty1212
  • misty1212
exactly!
anonymous
  • anonymous
Oh, and then my last root would look like this: 3(cos(196.6)+isin(196.6))
misty1212
  • misty1212
hmmm no
misty1212
  • misty1212
lets back up a second first angle was \(330\div 3=110\)
anonymous
  • anonymous
Yes
misty1212
  • misty1212
second one was \(\frac{330+360}{3}=230\)
anonymous
  • anonymous
Yes
misty1212
  • misty1212
third one add 360 again \[\frac{330+360+360}{3}\]
misty1212
  • misty1212
or more to your formula \[\frac{330+2\times 360}{3}\]
anonymous
  • anonymous
350
misty1212
  • misty1212
you keep going around the circle
misty1212
  • misty1212
right
misty1212
  • misty1212
not sure where the 196.6 came from
anonymous
  • anonymous
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!
misty1212
  • misty1212
\[\huge \color\magenta\heartsuit\] btw i hope it is now clear who easy this is
anonymous
  • anonymous
Crystal clear, thank you!
misty1212
  • misty1212
you're welcome dear

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