# Inverse tangent

1. Nov 9, 2016

### Karol

1. The problem statement, all variables and given/known data
Prove:
$$\tan^{-1}(-x)=-\tan^{-1}(x)$$

2. Relevant 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?

2. Nov 9, 2016

### Math_QED

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.

3. Nov 9, 2016

### PeroK

Why not use $y$ in these equations and then let $x = tan^{-1}(y)$?

4. Nov 9, 2016

### Karol

$$\left\{ \begin{array}{l} \tan(-y)=-\tan(y) \\ x=tan^{-1}(y) \end{array} \right.$$
$\tan(-y)=-\tan(y)$ refers to the function:

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

5. Nov 9, 2016

### Math_QED

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.

6. Nov 9, 2016

### PeroK

@Karol

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

$y = \tan(x)$

$-y = -\tan(x)$

Can you pick it up from there?

7. Nov 11, 2016

### Karol

I use $\tan(-x)=-\tan(x)$ to get:
$$-y = -\tan(x)=\tan(-x)~\rightarrow~-y=\tan(-x)$$
$$-y=\tan(-x)~\rightarrow~\tan^{-1}(-y)=-x$$
$$\left\{ \begin{array} {l} \tan^{-1}(-y)=-x \\ \tan^{-1}(y)=x \end{array} \right\}~\rightarrow~-\tan^{-1}(y)=\tan^{-1}(-y)$$
$$\rightarrow~-\tan^{-1}(x)=\tan^{-1}(-x)$$
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.

8. Nov 11, 2016

### PeroK

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.

9. Nov 11, 2016

### Karol

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

10. Nov 11, 2016