Question about arctan addition

[issacnewton]

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

 

[PeroK]

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 $y$?

 

[issacnewton]

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.

 

[PeroK]

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 $y$?

 

[issacnewton]

No, I am confused about the question.

 

[PeroK]

No, I am confused about the question.

Try some numbers. See what you get.

 

[issacnewton]

Ok, so the domain of $\arctan$ should be $(-\pi/2, \pi/2)$, right ? I will try some values

 

[PeroK]

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.

 

[issacnewton]

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 ?

 

[PeroK]

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 $\pm \pi$?

 

[issacnewton]

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.

 

[PeroK]

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 $\tan$.

 

[issacnewton]

So, is there no algebraic proof of this ?

 

[PeroK]

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.

 

[issacnewton]

Ok. Thanks

 

[wrobel]

If math analysis is allowed then one can consider derivative $\frac{\partial f}{\partial x}$

 

[haruspex]

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?

 

[wrobel]

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