Equivalent Arguments to the Tangent Function

[programmerBlack]

Hello,

Can someone tell me what equivalent angles can be passed into the Tangent function to get an equal value?

For example,
I have sin(x) = sin(pi - x) = sin(-pi - x), and cos(x) = cos(2pi - x) = cos(-2pi - x). I'm actually looking for the tangent versions of these. I cannot find them anywhere. Do they not exist?

Thanks in advance.

Edit:
What does is this technique called? Maybe some form of Trig identity? I'm not even sure what to search for.

 

[lev888]

Look at the graphs. "Draw" a straight horizontal line - wherever it intersects the graph, that's where the points of equal values are.
tan(x) = tan(x+n*pi) because pi is the period of tan(x). For any periodic function f(x) = f(x+n*period).

 

[programmerBlack]

Look at the graphs. "Draw" a straight horizontal line - wherever it intersects the graph, that's where the points of equal values are.
tan(x) = tan(x+n*pi) because pi is the period of tan(x). For any periodic function f(x) = f(x+n*period).

Oh wow. Thank you. I did not now it was that intuitive. I appreciate your response.

 

[Dr.Peterson]

I have sin(x) = sin(pi - x) = sin(-pi - x), and cos(x) = cos(2pi - x) = cos(-2pi - x). I'm actually looking for the tangent versions of these. I cannot find them anywhere. Do they not exist?

Your examples are related to reflections of the graphs of sine and cosine, which give rise to points within one cycle with the same value.

The tangent function doesn't have that kind of symmetry (reflection in a vertical line), so the only way you can get the same value is by its periodicity, a feature that all trig functions share.