# Inverse tangent

#### Karol

1. Homework Statement
Prove:
$$\tan^{-1}(-x)=-\tan^{-1}(x)$$

2. Homework Equations
Inverse tangent: $\tan(y)=x~\rightarrow~y=\tan^{-1}(x)$

3. The Attempt at a Solution
$$\tan(-x)=\frac{\sin(-x)}{\cos(-x)}=\frac{-\sin(x)}{\cos(x)}=-\tan(x)$$
I just change the unknown x to y:
$$\tan(-y)=\tan(y)$$
Now i have to translate it. we know:
$$\tan^{-1}(x)=\tan y~\rightarrow~-\tan^{-1}(x)=-\tan y$$
But:
$$\tan^{-1}(-x)~\rightarrow~\tan(y)=-x$$
Only from looking on the graph i can say $\tan(-y)=-x$ and finish but am i allowed to?

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#### Math_QED

Homework Helper
1. Homework Statement
Prove:
$$\tan^{-1}(-x)=-\tan^{-1}(x)$$
View attachment 108677

2. Homework Equations
Inverse tangent: $\tan(y)=x~\rightarrow~y=\tan^{-1}(x)$

3. The Attempt at a Solution
$$\tan(-x)=\frac{\sin(-x)}{\cos(-x)}=\frac{-\sin(x)}{\cos(x)}=-\tan(x)$$
I just change the unknown x to y:
$$\tan(-y)=\tan(y)$$
Now i have to translate it. we know:
$$\tan^{-1}(x)=\tan y~\rightarrow~-\tan^{-1}(x)=-\tan y$$
But:
$$\tan^{-1}(-x)~\rightarrow~\tan(y)=-x$$
Only from looking on the graph i can say $\tan(-y)=-x$ and finish but am i allowed to?
In a formal proof you cannot look at a graph to make conclusions. You only can use them to illustrate your proof but your proof should stand on its own.

#### PeroK

Homework Helper
Gold Member
2018 Award
3. The Attempt at a Solution
$$\tan(-x)=\frac{\sin(-x)}{\cos(-x)}=\frac{-\sin(x)}{\cos(x)}=-\tan(x)$$
Why not use $y$ in these equations and then let $x = tan^{-1}(y)$?

#### Karol

Why not use $y$ in these equations and then let $x = tan^{-1}(y)$?
$$\left\{ \begin{array}{l} \tan(-y)=-\tan(y) \\ x=tan^{-1}(y) \end{array} \right.$$
$\tan(-y)=-\tan(y)$ refers to the function:

While $x = tan^{-1}(y)~\rightarrow~\tan(x)=y$ refers to a different function:

#### Math_QED

Homework Helper
$$\left\{ \begin{array}{l} \tan(-y)=-\tan(y) \\ x=tan^{-1}(y) \end{array} \right.$$
$\tan(-y)=-\tan(y)$ refers to the function:
View attachment 108689
While $x = tan^{-1}(y)~\rightarrow~\tan(x)=y$ refers to a different function:
View attachment 108690
But you do know that $y = f(x) \iff x = f^{-1}(y)$, assuming that f has an inverse function?

A function and its inverse are related to each other, and that's exactly what you can use.

#### PeroK

Homework Helper
Gold Member
2018 Award
@Karol

$x = \tan^{-1}(y)$

$y = \tan(x)$

$-y = -\tan(x)$

Can you pick it up from there?

#### Karol

@Karol
$x = \tan^{-1}(y)$
$y = \tan(x)$
$-y = -\tan(x)$
Can you pick it up from there?
I use $\tan(-x)=-\tan(x)$ to get:
$$-y = -\tan(x)=\tan(-x)~\rightarrow~-y=\tan(-x)$$
$$-y=\tan(-x)~\rightarrow~\tan^{-1}(-y)=-x$$
$$\left\{ \begin{array} {l} \tan^{-1}(-y)=-x \\ \tan^{-1}(y)=x \end{array} \right\}~\rightarrow~-\tan^{-1}(y)=\tan^{-1}(-y)$$
$$\rightarrow~-\tan^{-1}(x)=\tan^{-1}(-x)$$
But you do know that $y = f(x) \iff x = f^{-1}(y)$, assuming that f has an inverse function?
A function and its inverse are related to each other
An inverse function is only the graph rotated -900, so why "assuming that f has an inverse function", there always is an inverse, no?
And why "A function and its inverse are related to each other"? they don't, they are completely different ones, also, no?
I mean if $y=\tan(x)$ then, if i rotate the graph and take from that y, again, $\tan(y)$ i will get something that has nothing in common with the original x.

#### PeroK

Homework Helper
Gold Member
2018 Award
An inverse function is only the graph rotated -900, so why "assuming that f has an inverse function", there always is an inverse, no?
And why "A function and its inverse are related to each other"? they don't, they are completely different ones, also, no?
I mean if $y=\tan(x)$ then, if i rotate the graph and take from that y, again, $\tan(y)$ i will get something that has nothing in common with the original x.
A function must be 1-1 to have an inverse. Normally, by restricting the domain of a function, you can make it 1-1 on the restricted domain. Like $\tan(x)$ on $(-\pi/2, + \pi/2)$.

I don't think you can say that a function and its inverse have nothing in common. They are very closely related.

#### Karol

Good, but what about my solution? is the rest (the beginning) O.K.?

#### PeroK

Homework Helper
Gold Member
2018 Award
Good, but what about my solution? is the rest (the beginning) O.K.?

#### Karol

Thank you PeroK and Math_QED

"Inverse tangent"

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