Free help is shit help

[Imperium]

Internet forums...

Sometimes they are useful because you can ask for help, but often the help you receive is of the handwaving variety.

All I need to know is why the arguments of a complex number in trigonometric form are added; proven from first principles.

Don't tell me to "go derive them from Euler's Identity." Stupid fucker, I am looking for an example!

 

[The Call Of Nature]

That's a bit general. Could you please explain the problem in more detail (and if possible in a simple English), so that I get an idea of what exactly you need?
I may perhaps then be able to provide an explanation and an example.


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Stop reading here if you're looking for a specific explanation. The following only explains to our less (or more?) fortunate friends what Euler's Identity is, how it finds a practical use and why complex numbers are involved.
==============================================

Generally speaking, Euler's Identity is a way to connect trigonometric functions and complex numbers. While complex numbers can normally not be used in trigonometric functions (for example sin 5i is not defined), Euler's identity converts any trigonometric function into a linear equation which you then can solve, regardless of whether it contains complex numbers or not.

The formula finds a practical use in alternating current network calculations when you try to determine the relation between voltage and current in a sinus-shaped current.
Another practical use is calculating the electrical impedance by means of a 2-part formula where the real part of impedance is the resistance and the imaginary (complex) part is the reactance.

 

[Imperium]

Show

(cos[a]cos - sin[a]sin) + j(sin[a]cos + cos[a]sin)

is equal to

cos(a+b) + jsin(a+b)

 

[Loktar]

Show

(cos[a]cos - sin[a]sin) + j(sin[a]cos + cos[a]sin)

is equal to

cos(a+b) + jsin(a+b)

x=42

 

[Blindgroping]

Show

(cos[a]cos - sin[a]sin) + j(sin[a]cos + cos[a]sin)

is equal to

cos(a+b) + jsin(a+b)

Um.... common distribution property?

 

[The Call Of Nature]

ah, I understand. You are supposed to use Euler's formula to prove Abu al Wafa Buzjani's angle sum and difference identities:

sin (A + B) = sinA cosB + cosA sinB
sin (A - B) = sinA cosB - cosA sinB
cos (A + B) = cosA cosB - sinA sinB
cos (A - B) = cosA cosB + sinA sinB

See what I mean? #1 and #3 are exactly your formula, only that in yours #3 gets multlpied with i


ok, let's give it a try:

[(cosA cosB) -(sinA sinB)] + i[(sinA cosB) + (cosA sinB)]

first we get rid of that i outside of the brackets by multiplying the stuff within the brackets with i

(cosA cosB) - (sinA sinB) + (i sinA cosB) + (i cosA sinB)

Looks already a bit friendlier and it looks familiar. We can sort this out, using the standard formulas, simply ignoring the i and treating it like any other number

(cosA +i sinA) (cosB + i sinB)

now (cosA + i sinA) looks awfully familiar, doesn't it? It is Euler's Identity* backwards.
We boldly use it to turn the problem into a nice exponential equation rather than this annoying sinus and cosinus stuff:
(Argh! My appologies, I must write it in words as I can't put letters a line higher here)

= (e to the power of iA) (e to the power of iB)

= e to the power of i(A+B)

and now we can apply Euler's identity the other way round again, and turning the complex exponentials into sin and cos again we get:

cos(A+B) + i sin(A+B)

q.e.d.


____________
* e to the power of ix = cos x + isin x

 

[The Call Of Nature]

*offers incense and freshly tapped beer on the altar of the board deities who set the maximum posting time up*
thanks, that post^ took me half an hour and I am not sure if I could do it again LOL - I am not that fit, intellectually, in the evening after only 4 hours sleep and 10 hours at work

no trolling for Jack today - I'm off to my bathtub

 

[The Call Of Nature]

[oops, wrong thread]

 

[dogbert]

What do you want for nothing? Rrrrrrrrrrrrrrrrrrrrrrrrrrrrrubber buscuit?

 

[Imperium]

(cosA cosB) - (sinA sinB) + (i sinA cosB) + (i cosA sinB)

Looks already a bit friendlier and it looks familiar. We can sort this out, using the standard formulas, simply ignoring the i and treating it like any other number

(cosA +i sinA) (cosB + i sinB)

I get lost with "using standard formulas", it has been years since I had to develop my Trigonometric identities.

I confess that I have not looked at this in depth, so I will have to review what "standard formulas" you are using to I can understand what you are doing to go from:

(cosA cosB) - (sinA sinB) + (i sinA cosB) + (i cosA sinB)

to:

(cosA +i sinA) (cosB + i sinB)

 

[The Call Of Nature]

I'm sorry, I wasn't online yesterday and saw this question only just now (5:37 am, on my way to work). Give me another 12 hours, please - I must work out in the country today without access to a PC plus I must try out the proof of the formula first. I just recalled it as a formula that used to turn up frequently at school and university, but since my last Maths lesson was more than 20 years ago, I am a little rusty.

I hope you aren't in a hurry

 

[The Call Of Nature]

Ok, I had to look this up myself as I normally just have to use the formulas and not reconstruct them from scratch.
There appear to be 2 ways to prove the addition rules of sin/cos. Since many characters can not be typed on a normal keyboard, I wrote them down and scanned them. I'm sorry that the files are so huge but I wanted a really good resolution.
I hope you can read my hand..

To keep it short, I did the proof only for one half of the formula. The other half goes analogue

We recapitulate: the problem was to proove that

(cosA + i sinA)(cosB + i sinB) =
= (cosA cosB) - (sinA sinB) + (i sinA cosB) + (i cosA sinB) =
= cos(a+b) + i sin(a+b)


1) Algebraic proof one which uses Euler's formula again and hence is not really that logical imho.
But maybe you can find the proof of this formula if you search the internet. Not being a native speaker of English I lack the special mathematical vocabulary and therefore don't know which keywords to search for. And I guess a German explanation won't be that helpful
algebra.tif (filesize: 4.15 MB)


2) a geometrical proof which imo is more logical. The algebraic one is interesting, though, as it brings the infamous i into the game. But as you can use i like any other number, it can easily be inserted in the geometrical proof as well.

addi.gif this is a small file, showing the triangles involved in the proof

geometry.tif the geometrical proof. (sorry, 33.1 MB huge )

I'm afraid that's really all info I could find. I hope it's sufficient.
I'll be afk all weekend. If there is a further problem, plase try an online search. You know better than I what the appropriate keywords would have to be as my (otherwise very reliable) dictionary doesn't cover special mathematical terms.
Good luck and have a nice weekend

 

[Imperium]

Thanks. I will look it over. It seems that the way to prove it is to move from the trigonometric form to the exponential form using Euler's Identity and then back again. Once in the exponential form, one can use properties of logarithms to add the arguments and then simply convert back.

Similar to taking an integro-differential equation, representative of the time domain, and transforming it using LaPlace transforms converting it into the frequency domain. The point being to avoid the nasty calculus and instead use algebra then finally convert back to the time domain using the Inverse LaPlace transform.

I had it in my mind for "first principles" I was thinking of doing it solely using trigonometric identities, but I am satisfied. I will look at your geometric proof too, that is always a fun read.

 

[Loktar]

I like you even less now that you're trying to pass yourself as a likeable math genious. Math sucks(and I don't like you) and that is why is why I decided to neg all your past(and future) posts in this thread.

 

[Imperium]

I like you even less now that you're trying to pass yourself as a likeable math genious. Math sucks(and I don't like you) and that is why is why I decided to neg all your past(and future) posts in this thread.

I am not a math genius. I leave many of the proofs to mathematicians while I stand on the mountain of their work.

BTW, I am assuming CoN is a math wiz, or at least has a background in math because he is using i for imaginary numbers. As an engineer I use j.

This certainly has been a great thread!

Anyway, neg away. It sounds like you need a good cheering up. :wink:

 

[Loktar]

I don't need any cheering up.

 

[The Call Of Nature]

BTW, I am assuming CoN is a math wiz, or at least has a background in math because he is using i for imaginary numbers. As an engineer I use j.

I'm no math wizz at all; quite the contrary, I'm afraid. I had to pass a maths test at university, though, as over here one needs a number of certificates (including maths) for being permitted to take the pre-diploma and diploma tests in biology.
When it comes to integro-differential equations you can totally count me out LOL. I've forgotten all about it in the last decades. I am surprised myself that I could reconstruct and find the formulas used in the above posts.
Oh, and btw, I'm female (and mid-40s, overweight and have a completely immortal (and sometimes immoral) sense of humour which not even my friend Loktar's negging can harm. If it's a comfort to you, he's been negging me for a week now, ruining my - until very recently - totally green karma list *grin*)

I had no idea engineers use j instead of i. I noticed you used it but thought it was for better legibility. Is there a particular reason why engineers use the j?


@ Loktar: maths can suck, I agree, but on the other hand it can also be fun and it has this perfect symmetry and order which is pretty somehow.
You know that sinus means bosom, do you? So you see that at least the sinus curve does have an esthetically pleasing practical use

What I like about maths is the intellectual challenge. I lose more fights than I win, but I am not someone who gives up that easily and it's so pleasing if after an eternal struggle, you finally get the upper hand and the problem surrenders.

When I can't sleep, I've been trying for years now to find a way to square a circle, but I couldn't recall the linear equation for a circle I had learned at highschool, 30 years ago. This Euler-business has set me on the right track, I think. Next time I can't sleep I'll try it out.

Indubitably, some people will now consider me a pervert, but I've been called much worse things over here and am quite confident that I'll survive it hpmrgreen:

 

[Loktar]

@ Loktar: maths can suck, I agree, but on the other hand it can also be fun and it has this perfect symmetry and order which is pretty somehow.
You know that sinus means bosom, do you? So you see that at least the sinus curve does have an esthetically pleasing practical use

So if told a woman " I want to bury my sinuses in her sinus." What is the mathematical probability I'd get slapped, tasered or pepper sprayed?

 

[The Call Of Nature]

50%, statistically.

With every woman you have 2 possibilities: she'll attack you or she won't.
And with every new woman the odds will be the same again, as you practically start from square one again.
So you'll always have a 50% chance that it works (or that it doesn't work)

Or in mathematical terms: chance = C = 1/(2!) = 1/(1*2) = 1/2 = 50%
(sorry, I just couldn't resist )

 

[Seph]

good advice is seldom cheap.