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

[Karol]

Homework Statement

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

Homework Equations

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

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?

 

[member 587159]

Homework Statement

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

Homework Equations

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

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]

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:

 

[member 587159]

$$\left\{ \begin{array}{l} \tan(-y)=-\tan(y) \\ x=tan^{-1}(y) \end{array} \right.$$
$\tan(-y)=-\tan(y)$ 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]

@Karol

What about:

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

$y = \tan(x)$

$-y = -\tan(x)$

Can you pick it up from there?

 

[Karol]

@Karol
What about:
$x = \tan^{-1}(y)$
$y = \tan(x)$
$-y = -\tan(x)$
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]

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]

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

Your solution is correct,

 

[Karol]

Thank you PeroK and Math_QED