Shouldn't be that hard considering that you can describe x and y as functions of r and theta too. Possibly some trigonometric identity which you should find in any larger math book.
Firstly, it isn't homework. Secondly, I already worked out the solution. Thirdly, neither of you know how to solve it. So, cut the crap and just admit you don't know.
imagine being filtered by algebra
I posted it because I know IQfy can't solve it. I'm filtering nearly all of IQfy with this.
2 years ago
Anonymous
> I'm filtering nearly all of IQfy with this.
Good job. You've filtered all of these anonymous users. You're a hero. >it isn't homework.
I am not buying this.
2 years ago
Anonymous
Firstly, you’re a Black person. Secondly, you’re a homosexual. Thirdly, suck my dick you dumb Black personhomosexual.
Now, OP, before someone goes off and embarrasses you in front of all of IQfy, I'll give you a chance to come clean. Are you absolutely sure that you already showed that they are indeed equal? Cause my hunch suggests that this problem is genuinely something that you contrived on your own. Think about it, there aren't any ... discrepancies in your work, are there?
I've already found they're equal. I'll give IQfy a few more days of squirming before I post it. FYI, this is a problem in Boas' Mathematical Methods for Physical Sciences. Also, the guy you responded to isn't OP -- I am.
Shouldn't be that hard considering that you can describe x and y as functions of r and theta too. Possibly some trigonometric identity which you should find in any larger math book.
not seeing any attempts by either of you. easy to say something is easy when you don't actually try it.
>not seeing any attempts by either of you
Because we aren't doing your homework, Jamal.
Frick off.
Firstly, it isn't homework. Secondly, I already worked out the solution. Thirdly, neither of you know how to solve it. So, cut the crap and just admit you don't know.
I posted it because I know IQfy can't solve it. I'm filtering nearly all of IQfy with this.
> I'm filtering nearly all of IQfy with this.
Good job. You've filtered all of these anonymous users. You're a hero.
>it isn't homework.
I am not buying this.
Firstly, you’re a Black person. Secondly, you’re a homosexual. Thirdly, suck my dick you dumb Black personhomosexual.
>not seeing anyone doing my hw for me
>curious
just get a list of identities, this is trivial
imagine being filtered by algebra
This board is pathetic. Nearly every instance of
>Not doing your homework
Translates to
>I can't solve it REEEEEEE
No, I don't think I will.
Now, OP, before someone goes off and embarrasses you in front of all of IQfy, I'll give you a chance to come clean. Are you absolutely sure that you already showed that they are indeed equal? Cause my hunch suggests that this problem is genuinely something that you contrived on your own. Think about it, there aren't any ... discrepancies in your work, are there?
I've already found they're equal. I'll give IQfy a few more days of squirming before I post it. FYI, this is a problem in Boas' Mathematical Methods for Physical Sciences. Also, the guy you responded to isn't OP -- I am.
>double angle formula, pythagoran theorem, etc
how's the implant situation?
my BALLS are erect
my scrotum is HARD
nice to hear *ZAAAAAP*
Are they equal? The top u negates when y negates, but in the bottom u only the arctan term negates, not the whole thing.
tangent is a periodic function
So what are you saying, they're equal mod pi or something?
imagine getting filtered by problem 1.
Then how come they ain't equal huh?
They are. I'll let you and others squirm for longer before I show the solution. Imagine trusting a computer more than pen and paper calculations.
>what is a principal value
Apply arctangent sum formula to
[eqn]A_{pm1}:= arctan({x over y}) - arctan({x pm 1over y}) = \arctanBig({x-(xpm 1)over y} Big/ (1+{xover y}{xpm1 over y})Big) = \
arctan({mp y over x^2 + y^2 pm x}) = arctan({mp yover r^2 pm x})[/eqn]
Apply sum formula again to [math]A_{1} + A_{-1}[/math]
[eqn]arctanBig( ({-yover r^2+x} + {yover r^2-x}) Big/ (1 + {y^2over(r^2-x)(r^2+x)})Big)=\
arctanBig( {2xy over r^4 - (x^2 - y^2) }Big)
[/eqn]
Insert polar formulas [math]x=rsintheta[/math], [math]y=rcostheta[/math] and use double angle identities.
[eqn]
arctanBig( {r^2 2sinthetacostheta over r^4 - r^2(sin^2theta - cos^2theta) }Big) = \
arctanBig( {sin2theta over r^2 + cos2theta }Big)
[/eqn]
Note the sign on the cos term. Putting it in gets equal results.
And OP reigns as the supreme fat homosexual over all fat homosexuals.
You got lost in the sauce. For example, look at the prefactors in the original image, moron. You don't have those.
Aha, the answer 2.2 because, iiiiii need a poo.