$\tan x= \cot (x+\phi)$ for some $\phi$

Question

Suppose we graphed the equation $ y = \tan x $. Is it possible to describe this graph with an equation of the form $ y = \cot (x + \phi) $, for some number $ \phi $? Why or why not?

Answer 1

Short Answer

No; there is not $ \phi $ such that $ y = \tan x $ can be expressed as $ y = \cot (x + \phi) $.

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Proof 1 - Algebraic & Rigorous

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Step 1: Find the relation between $ \phi $ and $ x $.

$ \tan x = \cot (x + \phi) $

$ \frac{\sin x}{\cos x} = \frac{\cos (x + \phi)}{\sin (x + \phi)} $

$ \cos (x + \phi)\cos x = \sin (x + \phi)\sin x $

$ \cos (x + \phi)\cos x - \sin (x + \phi)\sin x = 0 $

$ \cos ((x + \phi) + x) = 0 $

$ \cos (2x + \phi) = 0 $

$ 2x + \phi = \frac{\pi}{2} + k\pi $ (let $k$ be any integer)

$ \phi = -2x + \frac{\pi}{2} + k\pi $

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Step 2: Understand the relation between $ \phi $ and $ x $.

Sin $ \phi $ depends on $ x $, which is variable, there cannot be a constant $ \phi $ to satisfy $ \tan x = \cot (x + \phi) $.

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Proof - Graphically & Improper

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Step 1: Graph the functions $ y_0 = \tan x$ and $ y_1 = \cot x $.

(others are welcome to add images of some graphs)

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Step 2: Start shifting the function $ y_1 $ along the x-axis.

Note that you will never be able to superimpose $ y_1 $ over $ y_0 $.

Answer 2

If we had $\tan x=\cot(x+\phi)$, then the graph of $\tan x$ could be translated into the graph of $\cot x$, which it obviously cannot.

But compare the graphs below. I have translated them so that their asymptotes line up, but to bring them into coincidence we would also need to reflect about the $x$ axis. Of course, we can do that, which gives us the rather similar relation $$\tan x=-\cot(x-\frac{\pi}{2})$$

Answer 3

The function $$ f(x)=\tan x-\cot(x+\varphi) $$ is defined on an open set and everywhere differentiable (on its domain). Also $$ f'(x)=\frac{1}{\cos^2x}+\frac{1}{\sin^2(x+\varphi)}>0 $$ for every $x$ in the domain of $f$. Being a strictly increasing function on every interval included in the domain (which is a union of open intervals), $f$ is clearly not identically zero.

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Shorter proof: $\tan 0=\cot\varphi$ gives $\varphi=\pi/2+k_1\pi$; $\tan\pi/4=\cot(\pi/4+\varphi)$ gives $\cot(\pi/4+\varphi)=1$, so $\varphi=k_2\pi$. Contradiction.