Most elegant proof of Arctan(x) + Arctan(1/x)

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  • #1
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Hey guys, I was doing some questions from spivak when i noticed this integral he set up. To prove that the integral is [tex] \pi/2 [/tex] for all values of x, i needed to prove [tex] Arctan(x) + Arctan(\frac{1}{x}) = \pi/2 [/tex] for all values of x. Just wondering what is the most elegant proof of this?
 

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  • #2
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use the tangent addition formula:

[tex] \tan(u+v) = \frac{\tan u + \tan v}{1-\tan u \tan v} [/tex].

[tex] \frac{x+\frac{1}{x}}{0} = \tan \frac{\pi}{2} [/tex]

They are both undefined and approach [tex] \infty [/tex]
 
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  • #3
mathwonk
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jesus, just look at a triangle. it is obvious then that it is pi/2.
 
  • #4
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You have a right triangle. What do the two non 90 degree angles add up to? Oh yeah, that's right, 90 degrees. 90 degrees obviously = pi/2 radians. What more do you need? Arctan(x) finds one of the angles, Arctan(1/x) finds the other.

EDIT: Just realized mathwonk said the same thing.
 
  • #5
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Could someone provide a algebraic proof, i don't like pretty pictures.
 
  • #6
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You won't understand it if you don't know what Arctan(x) and Arctan(1/x) stand for!

Arctan(x) means the angle at which the the ratio of the opposite side length to the adjacent side length is x. Ok? So having 1/x simply switches the two sides, giving the other angle as the answer.

An algebraic proof may get you to believe that this is true, but you will have no idea why it's true. Saying that Arctan(x) + Arctan(1/x) = pi/2 is the same thing as saying the two non pi/2 radian angles add up to pi/2 radians.

Maybe it would be easier for you to think about degrees? Just substitute in 90 degrees for each.
 
  • #7
StatusX
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If you need to do it algebraically, note it's equivalent to show tan(x)tan(pi/2-x)=1.
 
  • #8
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I do not thinks the above equality is correct. Just take x = -1, we have arctan(x) = arctan(-1) = -Pi/4, and arctan(1/x) = arctan(1/-1) = arctan(-1) = -Pi/4. So, the sum of them is -Pi/2.

However, let consider the function f(x) = arctan(x) + arctan(1/x), x>0
On this interval, f(x) is differentiable and f'(x) = 0 for all x>0. Thus, f(x) = constant on this interval. In particularly, f(x) = f(1) = Pi/2 for all x>0

With the same function as above but defined with x<0, we also conclude that f(x) = -Pi/2 for all x<0
 
  • #9
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The proof using the picture is trivial, and i even generalised it before the post such that arctan(x/y) + arctan(y/x) = pi/2

but i need rigour.
 
  • #11
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arctg(x)+arctg(1/x)=sgn(x)*pi/2:confused:
 
  • #12
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then we have from trigonometry that:

[tex] Artanh(ix)=iartan(x) [/tex]

On the other hand...

[tex] 2artanh(x)=log(1-x)-log(1+x) [/tex]

[tex] 2artanh(ix)=log(1-ix)-log(1+ix) [/tex]

[tex] 2artanh(i/x)=log(1-i/x)-log(1+i/x) [/tex]

then [tex] artanh(ix)+artanh(i/x)=i(artan(x)+artan(1/x)) [/tex]

taking the sum of all the logs you have..

[tex] -log(1+ix)-log(1+i/x)+log(1-ix)+log(1-i/x) [/tex]

[tex] -log(2i/x)+log(-2i/x)=log(-1)=i\pi [/tex]

- i have taken the first "branch" of log ..(the angle goes from 0 to 360 degrees), the factor "2" comes from the definition of artanh(x) in the form of log (log in basis e).. :redface:
 
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  • #13
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take the derivative of the lhs you will find it is equal to 0 for all x. Then you just plug a value into x to find the constant in particular domain. In addition, f(0) is undefined. I hope this is simple enough.
 
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  • #14
Gib Z
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I do not thinks the above equality is correct. Just take x = -1, we have arctan(x) = arctan(-1) = -Pi/4, and arctan(1/x) = arctan(1/-1) = arctan(-1) = -Pi/4. So, the sum of them is -Pi/2.
arc tan has to be set a domain, which you are out of.

And to whoever asked the question, maybe arctan(1/x) will look better for you if you call it arccot x.

arc tan x + arc cot x = pi/2 is just another way of expressing the supplementary relationship tan [(pi/2) - x] = cot x

You could go by an unnecessary method of proof involving calculus...
Let f(x) = Arctan(x) + Arctan(1/x)

We know the derivative of Arctan(x) = 1/(1+x^2).If you didnt already know that, tell me and ill post my proof. Anyway, using that derivative for arc tan, and letting u=1/x, then using the chain rule, we evenutally get my f(x)'s derivative is equal to zero.

Since the gradient is zero, the answer is constant, unchanging. That means we can just sub in any number and get our value for all x. Easiest to use x=1

Arctan(1) = pi/4

therefore Arctan(x) + Arctan(1/x) = pi/2 . Not so elegant, but works.
 
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  • #15
Gib Z
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Ugh ive been trying to like 20 mins to use tex, but it just wont work for me..zz..

Edit: O god damn crap it, I just read the post before me saying exactly da same...kill me...
 
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  • #16
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"arc tan has to be set a domain, which you are out of."

What domain?Are you sure about this?
 
  • #17
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The way to prove it by Gib 7 is simple but effective. Thanks
 
  • #18
Gib Z
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Of course arc tan has to be set a domain, otherwise there is an infinite number of solutions for say, arc tan 1. Theres 45 degrees, then 225, 405, 595 so on so forth. With more than 1 solution, it is no longer a function and can not be differentiated.

And, Its Gib "Z" not "7" lol
 
  • #19
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didn't i do the same thing?
 
  • #20
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but -1 belongs to that domain and result is between -pi/2 and pi/2;

there is nothing wrong with arctg(-1)+arctg(1/-1)=-pi/2 !

thus,arctg(x)+arctg(1/x)=sgn(x)*pi/2
 
  • #21
Gib Z
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lol ok my bad i didnt read your posts well enough. my mistake..just randomly, how come i see it says 1 post for milos, when hes done 2 on this page lol?
 
  • #22
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use the identity,
[tex]\arctan{x}+\arctan{y}=\arctan{\frac{x+y}{1-xy}}[/tex]

which is easily derived from the tan addition formula
 
  • #23
Gib Z
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Thats what would seem to be the solution except your denominator equals 0, close but no cigar.
 
  • #24
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Let us say there is a triangle ABC with <ABC a right angle.

arctan(AB/BC)=<ACB
arccot(AB/BC)=<CAB

Knowing [tex]<ACB+<CAB+<BAC=\pi[/tex], it would be really obvious.
 

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