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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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    @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 :-)2012-01-11

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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)}. $


enter image description here


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: Understood. (Nice use of colors in the text and on the figure, by the way.)2012-01-11