De Moivre's Theorem

gimme5minutes

Well Ive been studying complex numbers and its been grand so far, all makes sense. However I'm now looking at a question that says "State De Moivre's theorem and use it to show that: Sin 3Θ = 3Sin Θ - 4Sin^3 Θ".

Well I know De Moivre's theorem is [r(Cos Θ + jSin Θ]^n = r^n[Cos nΘ + jSin nΘ] and how to apply it generally but I've never seen it applied to a function like the above which doesn't even contain a complex number from what I can see. Anybody able to give me some help here so I can make a stab at it?

LeixlipRed

The key to this is there's another way to write the sin function involving complex numbers. Have a google

FoxT

De Moivre gets you started, but you have a bit of mullocking to do after that...

LeixlipRed

The OP will not be served by people just posting solutions. Can we please have more pointing in the right direction and less giving the whole answer.

gimme5minutes

I googled for an hour and a half last night before I posted that and Ive been googling another 45 minutes this morning, to no avail. I know z = a + bi can be written as z = | z | [ cos (x) + i sin (x) ] but I can't see any examples on how this relates back to the example I posted above. I actually haven't found one similar question in a similar format to Sin 3Θ = 3Sin Θ - 4Sin^3 Θ, and how an equation in that form can be rewritten as a complex number, during my searches. I would appreciate a bit more help here as I can't even see where to start.

seandoiler

how about you start with (Cos(theta) + i Sin(theta))^3...now what does this give using deMoivre?

now think about expanding the left hand side and comparing

FoxT

Leixlip Red,
That was petty & annoying.

I have seen no rule or sticky on this forum that states solutions cannot be posted. If you want to apply that rule to this forum then let people know. Making rules up as you go along is not the role of a mod.

Also, It is not for you to decide whether the solution will benefit the OP or not. Every math student gets stuck from time to time, I believe that studying solutions is an important part of learning about the subject. (But I dont delete posts of people who disagree with my point of view)

FoxT

LeixlipRed

FoxT wrote: »

Leixlip Red,
That was petty & annoying.

I have seen no rule or sticky on this forum that states solutions cannot be posted. If you want to apply that rule to this forum then let people know. Making rules up as you go along is not the role of a mod.

Also, It is not for you to decide whether the solution will benefit the OP or not. Every math student gets stuck from time to time, I believe that studying solutions is an important part of learning about the subject. (But I dont delete posts of people who disagree with my point of view)

FoxT

We have a huge issue with regards to students coming on here looking for soutions to questions which they can in turn learn off for an exam. Students I know in person have done it on here before. It has been discussed on the moderators forum (a forum you do not have access to) to amend the charter to explicitly state that posters are not allowed to give full solutions to assignment/exam related questions. I have been to busy to come up with a segment to be inserted into the charter but plan to do so soon. As for your point about solutions being useful, that may be true, but only after the student has exhausted all other methods of attempting the question themselves. Thank you for your time.

EDIT: Also, your post has not been deleted. If the Op eventually works it out themselves I'll revert back to your original post.

LeixlipRed

Just a note to all users that the charter has now been amended. Hope that clears up any confusion.

FoxT

Thank you for updating the forum charter, makes the issue clear to all.

-FoxT

f1dan

[r(Cos Θ + jSin Θ]^n = r^n[Cos nΘ + iSin nΘ]

Hint: Binomial theorem

(CosΘ + iSinΘ)^n =
n
E (n)(CosΘ^n-k)(SinΘ^k)
(k)
k=0

It's badly typed but you should get the idea

squidge1

(cosƟ + isinƟ)^n gives you what according to De Moivre and gives you what according to plain expansion.

Hint: (x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3