Question about arctan addition

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I think this is because the domain of the arctan function is ##(-\pi/2, \pi/2)##, so if the value exceeds this range, we add or subtract ##\pi## to get a value within the range.
  • #1
issacnewton
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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
 
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  • #2
IssacNewton said:
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
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
IssacNewton said:
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:

PeroK said:
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
No, I am confused about the question.
 
  • #6
IssacNewton said:
No, I am confused about the question.

Try some numbers. See what you get.
 
  • #7
Ok, so the domain of ##\arctan## should be ##(-\pi/2, \pi/2)##, right ? I will try some values
 
  • #8
IssacNewton said:
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
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
IssacNewton said:
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
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
IssacNewton said:
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
So, is there no algebraic proof of this ?
 
  • #14
IssacNewton said:
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
Ok. Thanks
 
  • #16
If math analysis is allowed then one can consider derivative ##\frac{\partial f}{\partial x}##
 
  • #17
IssacNewton said:
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.
wrobel said:
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
haruspex said:
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
 

FAQ: Question about arctan addition

What is arctan addition?

Arctan addition is a mathematical operation that involves adding two arctan values together. Arctan, also known as inverse tangent, is the inverse function of tangent and is used to find the angle in a right triangle given the ratio of its sides.

How do you add two arctan values?

To add two arctan values, you can use the formula: arctan(a) + arctan(b) = arctan((a + b) / (1 - ab)). This formula is derived from the tangent addition formula and can be used to find the arctan value of the sum of two angles.

What is the range of arctan addition?

The range of arctan addition is [-π/2, π/2], which is the same as the range of the arctan function. This means that the result of adding two arctan values will always be within this range.

Can arctan addition be used for non-right triangles?

No, arctan addition can only be used for right triangles. In non-right triangles, the angles are not related by trigonometric functions and therefore, the formula for arctan addition cannot be applied.

How is arctan addition used in real life?

Arctan addition is used in various fields such as engineering, physics, and navigation. It can be used to calculate the direction and angle of a moving object, determine the slope of a surface, and solve problems involving right triangles.

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