Question

Find the cube roots of 27(cos 330° + i sin 330°).

Answer 1

Lol....u Know De-Morvies Theorem

Answer 2

I think it is (costheta + i sintheta)^n = (cosn theta + i sin n theta)

Answer 3

(cosn theta + i sin n theta) = e^(i theta)

Answer 4

27(cos 330° + i sin 330°) = 27 e^(i 11pie/6)

Answer 5

now find its Cube Root....)

Answer 6

I think I did this wrong. 44796.22502i?

Answer 7

@Yahoo!

Answer 8

Your Euclids theorem is wrong I think.. @Yahoo!

Answer 9

\[\large \cos(\theta) + i \sin(\theta) = e^{i \theta}\]

According to Euler Therorem..

Answer 10

This can be solved by using De-Moivre's theorem only..

Answer 11

Tell me, according to that theorem, what will you get after this??
\[\large (27(\cos(330) + i \sin(330))^{\frac{1}{3}} = ??\]

Answer 12

oh...Sorry....i understood this question Differently..

Answer 13

So what do i do?

Answer 14

Use here De-Moivre's theorem..

Here n = 1/3..

Answer 15

I think you have written formula for the theorem above..

Use that @Itel

Answer 16

@Yahoo! @waterineyes It's Euler's Formula :p

Answer 17

More precisely, that is one of Euler's Identity.. @parthkohli ..

Answer 18

I keep getting 26.58980933-4.688500797i

Answer 19

\[\huge 27(\cos 330^o+i \sin 330^o)=27e^{\frac{11\pi}{6}i}\]Taking the cube root gives us,\[\huge \left(27e^{\frac{11\pi}{6}i}\right)^{1/3}\quad=\quad 3e^{\frac{11\pi}{18}i}\]This angle converted to degrees is,\[\huge \frac{11\pi}{18}\cdot \frac{180}{\pi} \quad = \quad 110^o\]Yahoo had the right idea. You can certainly do it this way if you want :O Using that theorem is probably faster though :)

Answer 20

\[\large 3(\cos 110^o+i \sin 110^o)\]

Answer 21

It keeps popping up as [Math processing Error] in your response. So I can't really figure out what it is saying. Sorry. I'm just completely confused on this one. I tried using the theorem and I got this way off the wall answer. I tried some other relational thing I found on google and that gave me an even more bizarre answer.

Answer 22

Hey when you get [Math processing Error] then reload or refresh your page..

You will see everything then..

Answer 23

\[\large [27(\cos(330) + i \cdot \sin(330))]^{\frac{1}{3}} = (27)^{\frac{1}{3}}(\cos(\frac{1}{3} \times 330) + i \cdot \sin(\frac{1}{3} \times 330))\]

Answer 24

@waterineyes Let me get it straight :)
\[\textbf{EULER'S IDENTITY} \ \ \ \ e^{\pi i} + 1 = 0\]\[\textbf{EULER'S FORMULA} \ \ \ \ e^{i \phi} = \cos(\phi) + i\sin(\phi)\]

Answer 25

Finally you will get as @zepdrix said:

\[\implies \color{blue}{3 [\cos(110) + i \cdot \sin(110)]}\]

Answer 26

Euler gave two Identities :

\[e^{i \theta} = \cos(\theta) + i \sin(\theta)\]
\[e^{-i \theta} = \cos(\theta) - i \sin(\theta)\]

If I am not wrong then..

Answer 27

Yes, the second is just a derivative of the second. :)

Answer 28

What??

Answer 29

Second is just derivative of second??

Reflexive Property?? Ha ha ha..

Answer 30

Dusra wala formula pehle se derive hua hai kaakey :D

Answer 31

It's pretty easy to do so:\[e^{-i\phi} = e^{i\times(-\phi)} = \cos(-\phi) + i\sin(-\phi) = \cos(\phi) - i\sin(\phi). \]Knowing that \(\sin(-x) = -\sin(x)\) and \(\cos(-x) = \cos(x).\)

Answer 32

Yes, the second is just a derivative of the second. :)

The second time you have written second should be first I think.. :P

Answer 33

The second is just a derivative of the first*

Answer 34

Okay..

Yes you are right..

and thanks for clearing me straight on Euler's Identity and Formula..

Answer 35

It's not the derivative of the first. It's just the identity for -theta. Given that Cosine is even, Sine is an odd function, the negatives move around nicely.

If it was the derivative it would look something like...\[\large ie^{i \theta}=-\sin \theta+i \cos \theta\]
I'm not sure what that formula would tell us :D heh

Answer 36

Parthkohli, just meant to say that second one can be easily derived from First formula..

He is not saying about Differentiation or Derivative..

Answer 37

Oh my mistake ^^