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Inverse tangent

  • Thread starter Karol
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1. Homework Statement
Prove:
$$\tan^{-1}(-x)=-\tan^{-1}(x)$$
Snap1.jpg


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?
 

Math_QED

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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

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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)##?
 
1,380
22
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:
Snap1.jpg

While ##x = tan^{-1}(y)~\rightarrow~\tan(x)=y## refers to a different function:
Snap1.jpg
 

Math_QED

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

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@Karol

What about:

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

##y = \tan(x)##

##-y = -\tan(x)##

Can you pick it up from there?
 
1,380
22
@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

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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.
 
1,380
22
Good, but what about my solution? is the rest (the beginning) O.K.?
 
1,380
22
Thank you PeroK and Math_QED
 

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