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

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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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1 Answer

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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. $$ \color{blue}{ \tan(x)}\longrightarrow\color{maroon}{\tan(-x)} =\color{maroon}{\sin(-x)\over \cos(-x)}\longrightarrow \color{darkgreen}{{\sin\bigl(\textstyle{-(x-{\pi\over2}})\bigr)\over \cos\bigl(\textstyle{-(x-{\pi\over2}})\bigr)}}= \color{darkgreen}{{\sin(\textstyle{\pi\over2}-x)\over \cos(\textstyle{\pi\over2}-x)}}= \color{darkgreen}{{\cos( x)\over \sin( x)}}=\color{darkgreen}{\cot (x)}. $$

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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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