The graph of $\cot$ is the image of the graph of $\tan$ by a simple transformation

How can I justify that the graph of the function cotangent : $\cot$ is the image of the graph of the function $\tan$ by a simple transformation.

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How can you? Easy: think! – Did Jan 11 '12 at 11:00
In particular think of how the two are defined in terms of $\sin$ and $\cos$. – David Mitra Jan 11 '12 at 11:20
I think my question is more : does the function x|-> 1/x refer to a simple transformation of the plan ? I'm wondering about the rotation. – user20010 Jan 11 '12 at 11:27
Not rotation, but inversion. You won't get it from just translations. But you could do it with a reflection through $y$ followed by a translation. – David Mitra Jan 11 '12 at 12:20
@DidierPiau : It turns out inversion is no longer in the curriculum which is taught in 12th (that's is my grade) in my country. Thus, it was pretty useful to ask for it here :-) – user20010 Jan 11 '12 at 12:23

I think inversion is simple enough: $$\color{blue}{ \tan(x)}\longrightarrow\color{darkgreen}{{1\over \tan x}}=\color{darkgreen}{\cot (x)}.$$

If not, then you'd need two transformations: reflection through the $y$-axis followed by a horizontal shift. $$\color{blue}{ \tan(x)}\longrightarrow\color{maroon}{\tan(-x)} =\color{maroon}{\sin(-x)\over \cos(-x)}\longrightarrow \color{darkgreen}{{\sin\bigl(\textstyle{-(x-{\pi\over2}})\bigr)\over \cos\bigl(\textstyle{-(x-{\pi\over2}})\bigr)}}= \color{darkgreen}{{\sin(\textstyle{\pi\over2}-x)\over \cos(\textstyle{\pi\over2}-x)}}= \color{darkgreen}{{\cos( x)\over \sin( x)}}=\color{darkgreen}{\cot (x)}.$$

See Didier Piau's comment below. If one considers a reflection through an arbitrary line as a simple transformation, then only one transformation is necessary.

In the last paragraph above, I was considering "simple transformation" as one of:

$\ \ \ \bullet\$reflection through the x-axis

$\ \ \ \bullet\$reflection through the y-axis

$\ \ \ \bullet\$horizontal/vertical shifts

$\ \ \ \bullet\$horizontal/vertical scaling by positive factors.

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David: you'd need two transformations... Consider the symmetry with respect to the line of equation $x=\frac{\pi}4$. – Did Jan 11 '12 at 13:30
@Didier Piau Of course, you're right; but I was considering "simple transformation" as one of: reflection through the $x$-axis, reflection through the $y$-axis, horizontal/vertical shifts, and horizontal/vertical scaling. At least, here in the US, those are the transformations studied in the usual precalculus course. – David Mitra Jan 11 '12 at 13:48
David: Understood. (Nice use of colors in the text and on the figure, by the way.) – Did Jan 11 '12 at 14:18