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Question about arctan addition

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  • #1
904
17

Homework Statement:

Prove the following identity . If ## f(x) = \arctan(x) + \arctan(y) ## then prove that
$$f(x) =
\begin{cases}
\arctan\bigg( \frac{x+y}{1 - xy} \bigg), \qquad xy < 1 \\
\pi + \arctan\bigg( \frac{x+y}{1 - xy} \bigg), \quad x>0, y > 0, xy > 1 \\
-\pi + \arctan\bigg( \frac{x+y}{1 - xy} \bigg), \quad x < 0, y < 0, xy > 1
\end{cases} $$

Relevant Equations:

Equations of inverse tangent functions
Now following is my attempt for the very first one. Let ##\arctan(x) = \alpha## and ## \arctan(y) = \beta ##. So, I have ## x = \tan(\alpha)## and ##y = \tan(\beta)##. Since the domain is restricted to define inverse function, I have that ##\alpha \in (-\pi/2, \pi/2) ## and ##\beta \in (-\pi/2, \pi/2)##. Now, I use the following identity for addition of angles

$$ \tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta) }{ 1- \tan(\alpha) \tan(\beta) } $$

On the left hand side, we have ##\alpha + \beta ## as the argument for the ##tan## function. But since the domain of the tan function is restricted, we must have that ##\alpha + \beta \in (-\pi/2, \pi/2) ##. This further restricts the values which ##\alpha## and ##\beta## can take. So, we have ##\alpha \in (-\pi/4, \pi/4) ## and ##\beta \in (-\pi/4, \pi/4)##. This means the following

$$ - \frac{\pi}{4} < \alpha < \frac{\pi}{4} $$
$$ - \frac{\pi}{4} < \beta< \frac{\pi}{4} $$

Now, I use the fact that ##\tan## is an increasing function. So, I get ## - 1 < \tan(\alpha) < 1 ## and ## -1 < \tan(\beta) < 1 ##. Which means that ##x \in (-1,1) ## and ##y \in (-1,1) ##. From this, its easy to prove that ## xy < 1## and plugging for x and y, we get the required first identity.

$$ \alpha + \beta = \arctan \Big( \frac{x+y}{1 - xy} \Big) $$
$$ \arctan(x) + \arctan(y) = \arctan \Big( \frac{x+y}{1 - xy} \Big) $$

Is my reasoning right ? For the other two identities, I have no clue how to proceed.
Thanks
 

Answers and Replies

  • #2
PeroK
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Homework Statement:: Prove the following identity . If ## f(x) = \arctan(x) + \arctan(y) ## then prove that
$$f(x) =
\begin{cases}
\arctan\bigg( \frac{x+y}{1 - xy} \bigg), \qquad xy < 1 \\
\pi + \arctan\bigg( \frac{x+y}{1 - xy} \bigg), \quad x>0, y > 0, xy > 1 \\
-\pi + \arctan\bigg( \frac{x+y}{1 - xy} \bigg), \quad x < 0, y < 0, xy > 1
\end{cases} $$
Relevant Equations:: Equations of inverse tangent functions

Now following is my attempt for the very first one. Let ##\arctan(x) = \alpha## and ## \arctan(y) = \beta ##. So, I have ## x = \tan(\alpha)## and ##y = \tan(\beta)##. Since the domain is restricted to define inverse function, I have that ##\alpha \in (-\pi/2, \pi/2) ## and ##\beta \in (-\pi/2, \pi/2)##. Now, I use the following identity for addition of angles

$$ \tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta) }{ 1- \tan(\alpha) \tan(\beta) } $$

On the left hand side, we have ##\alpha + \beta ## as the argument for the ##tan## function. But since the domain of the tan function is restricted, we must have that ##\alpha + \beta \in (-\pi/2, \pi/2) ##. This further restricts the values which ##\alpha## and ##\beta## can take. So, we have ##\alpha \in (-\pi/4, \pi/4) ## and ##\beta \in (-\pi/4, \pi/4)##. This means the following

$$ - \frac{\pi}{4} < \alpha < \frac{\pi}{4} $$
$$ - \frac{\pi}{4} < \beta< \frac{\pi}{4} $$

Now, I use the fact that ##\tan## is an increasing function. So, I get ## - 1 < \tan(\alpha) < 1 ## and ## -1 < \tan(\beta) < 1 ##. Which means that ##x \in (-1,1) ## and ##y \in (-1,1) ##. From this, its easy to prove that ## xy < 1## and plugging for x and y, we get the required first identity.

$$ \alpha + \beta = \arctan \Big( \frac{x+y}{1 - xy} \Big) $$
$$ \arctan(x) + \arctan(y) = \arctan \Big( \frac{x+y}{1 - xy} \Big) $$

Is my reasoning right ? For the other two identities, I have no clue how to proceed.
Thanks
I don't see where you used ##xy < 1## here.

Do you understand why ##f(x)## is not always equal to ##g(x) = \arctan(\frac{x + y}{1 - xy})##, depending on ##x## and ##y##?
 
  • #3
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Well ##x## and ##y## are in ##\mathbb{R}## and since I reasoned that ##x, y \in (-1,1)##, we can take various cases of ##x## and ##y## and then prove that ##xy < 1##. I thought we are supposed to prove this.
 
  • #4
PeroK
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Well ##x## and ##y## are in ##\mathbb{R}## and since I reasoned that ##x, y \in (-1,1)##, we can take various cases of ##x## and ##y## and then prove that ##xy < 1##. I thought we are supposed to prove this.
If ##xy < 1##, you could have ##x = \frac 1 3, \ ##y = 2##.

What about my main question:

Do you understand why ##f(x)## is not always equal to ##g(x) = \arctan(\frac{x + y}{1 - xy})##, depending on ##x## and ##y##?
 
  • #5
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No, I am confused about the question.
 
  • #6
PeroK
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No, I am confused about the question.
Try some numbers. See what you get.
 
  • #7
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Ok, so the domain of ##\arctan## should be ##(-\pi/2, \pi/2)##, right ? I will try some values
 
  • #8
PeroK
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Ok, so the domain of ##\arctan## should be ##(-\pi/2, \pi/2)##, right ? I will try some values
The domain of ##\arctan## is ##\mathbb R##. The range is ##(-\pi/2, \pi/2)##.

It's a tricky question.
 
  • #9
904
17
I tried few values of ##x,y## so that ##xy < 1## on WolframAlpha and yes in this case, we have that
$$ \arctan(x) + \arctan(y) = \arctan\Big( \frac{x+y}{1-xy} \Big) $$

So, how do I proceed proving this ? How would my argument change ?
 
  • #10
PeroK
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I tried few values of ##x,y## so that ##xy < 1## on WolframAlpha and yes in this case, we have that
$$ \arctan(x) + \arctan(y) = \arctan\Big( \frac{x+y}{1-xy} \Big) $$

So, how do I proceed proving this ? How would my argument change ?
What I meant was to try to find values of ##x, y## where this is not true and see why. The confusing thing, surely, is why that equation is not true for all ##x, y##. Why do you sometimes have to add ##\pm \pi##?
 
  • #11
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So, for ##x=10## and ##y = 1/3##, I get ##\arctan(x) + \arctan(y) = 1.793##, which is greater than ##\pi/2## , so this is is beyond the range of arctan function and ##\arctan((x+y)/(1-xy)) = -1.35## and this is within the range of arctan function.
 
  • #12
PeroK
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So, for ##x=10## and ##y = 1/3##, I get ##\arctan(x) + \arctan(y) = 1.793##, which is greater than ##\pi/2## , so this is is beyond the range of arctan function and ##\arctan((x+y)/(1-xy)) = -1.35## and this is within the range of arctan function.
... and the difference is ##1.793 - (-1.35) = 3.14 = \pi##.

What I would do first is draw the graph of ##\tan## and show various possibilities for ##\alpha, \beta## and ##\alpha + \beta##.

Because of the nature of the ##\arctan## function, this isn't going to come out purely algebraically. You're going to need the graph of ##\tan##.
 
  • #13
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So, is there no algebraic proof of this ?
 
  • #14
PeroK
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So, is there no algebraic proof of this ?
I said "purely" algebraically. The problem is knowing when to add ##\pm \pi##. That's not going to come out very easily without using the graph.
 
  • #15
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Ok. Thanks
 
  • #16
If math analysis is allowed then one can consider derivative ##\frac{\partial f}{\partial x}##
 
  • #17
haruspex
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is there no algebraic proof of this ?
The difficulty is that the arctan function is discontinuous. You know that ##\arctan(\frac{x+y}{1-xy})## will produce the right value modulo ##\pi##. It remains to show that for the second case it produces a value in the range ##(-3\pi/2, -\pi/2)##, etc.
If math analysis is allowed then one can consider derivative ##\frac{\partial f}{\partial x}##
How does that help with the question asked by the OP, namely, dealing with the different ranges?
 
  • #18
How does that help with the question asked by the OP, namely, dealing with the different ranges?
Let us regard ##y## as a parameter. We have two functions
$$f(x)=\arctan x+\arctan y,\quad g(x)=\arctan\Big(\frac{x+y}{1-xy}\Big).$$
Assume for example that ##y>0,\quad x>1/y##.
We obviously have ##g'(x)=f'(x)##. And these derivatives are continuous for ##x>1/y##. Thus for such ##x## it follows that ##g(x)=f(x)+const.## The constant may depend on the parameter ##y##.
Passing to a limit as ##x\to 1/y+## we get $$-\pi/2=\arctan(1/y)+\arctan y+const.$$
On the other hand
$$\frac{d}{dy}(\arctan(1/y)+\arctan y)=0$$ and thus (chek for ##y=1##) we get
$$\arctan(1/y)+\arctan y=\pi/2\Longrightarrow const=-\pi$$
That's all
 

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