Please help me solve this problem.

if a=cos2alpha + isin2alpha

b=cos2beta + isin2beta

prove that,

a-b / a+b = itan (alpha-beta)

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- August 19th 2012, 07:53 PMok59[Solved] De Moivre's Theorem
Please help me solve this problem.

if a=cos2alpha + isin2alpha

b=cos2beta + isin2beta

prove that,

a-b / a+b = itan (alpha-beta) - August 19th 2012, 08:15 PMrichard1234Re: De Moivre's Theorem
Assuming "2alpha" is , then

Then, . Can you take it from there? - August 19th 2012, 08:23 PMok59Re: De Moivre's Theorem
sorry i dont get it ( still learning )

how do i suppose to get tan from it.

do you mean (cos alpha + isin alpha)^2 - August 19th 2012, 08:59 PMrichard1234Re: De Moivre's Theorem
Yes, it's also equal to by de Moivre's. I don't know if that helps though. Expanding doesn't yield anything useful.

It could also help to play around with the RHS, . Note that this is a purely imaginary number. You can try to reduce the LHS as or something like that. You'll most probably need your trigonometric identities. - August 19th 2012, 09:08 PMok59Re: De Moivre's Theorem
hmm.. this problem is so freaking weird . I am about to give up ( it has been 2 days).

- August 19th 2012, 09:36 PMrichard1234Re: De Moivre's Theorem
Yeah, it's an interesting problem. I haven't solved it yet...there seem to be several possible ways to the solution though. Just try a bunch of methods, see if one of them works. Maybe another user has a nice, elegant solution that doesn't require a lot of brute force (I don't want to brute force algebra/trig identities right now...). There might even be a geometric solution involving two points on a circle and finding the tangent of the angle in between.

- August 19th 2012, 10:22 PMDevenoRe: De Moivre's Theorem
let's write, for the time being, cos(2α) = u, sin(2α) = v, cos(2β) = y, sin(2β) = z.

then:

now we can use some trig to simplify.

note that u^{2}+v^{2}= y^{2}+z^{2}= 1. so our messy fraction becomes:

now

and .

so what we have is:

- August 20th 2012, 06:36 AMok59Re: De Moivre's Theorem
Man ! Thank so very much Deveno.

You made it so simple to understand mate. You can't imagine the time i've spent on this problem.

Thank You again for your help.

Have a nice day !

edit: Thank you to Richard1234 too. I have solved it by your way too. Now I have two solutions.(Cool) - August 20th 2012, 06:45 AMSorobanRe: De Moivre's Theorem (Part 1)
Hello, ok59!

Out of curiosity, I brute-forced it . . . whew!

We need the Sum-to-Product Identities:

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- August 20th 2012, 12:24 PMok59Re: De Moivre's Theorem (Part 1)
Thanks Mr. Soroban for another method . Now I understand that the trick is to divide and multiply with the conjugate of denominator to get the result. I appreciate the efforts of the members of this board.

- September 12th 2012, 10:39 PMmanoj9585Re: De Moivre's Theorem
De Moivre's Theorem is a relatively simple formula for calculating powers of complex numbers.De Moivre's formula states that for any real number x and any integer n, (cosx + isinx)n = cos(nx) + isin(nx).(r cisθ)^n = r^n cis(nθ) here n is an integer.

free algebra word problems - September 13th 2012, 02:04 AMjohnsomeoneRe: [Solved] De Moivre's Theorem
Here's another (equivalent) way: I'll begin with a general calculation that I'll then apply to this specific problem.

Suppose , and (i.e. on the unit circle except the leftmost point).

So let , where is not an odd multiple of .

Then

.

With that in hand, the result follows quickly. Let , . Let . (Note that , so .)

Let . Then , and obviously .

Notice that from , it follows that , so that , and hence so can apply the above derivation with this .

Now consider . Have:

as desired.

Thus the claim is established.

Notice that all the "bad" cases amounted to , because then , so , so .

That would've made , and also would've maded , which is where is undefined.