Does $\tan(x)$ and $\cot(x)$ has symmetry axis? (like e.g $\cos(x)$ at $\pi k$ for $k \in \mathbb{Z}$), I tried think in the direction that $\sin(x)/\cos(x) = \tan(x)$ and both of them have symmetry axis but I couldn't quite get around the solution.
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$\begingroup$ First, velcome to our site! What did you try to do? tell us! $\endgroup$kjetil b halvorsen– kjetil b halvorsen2014-08-15 09:16:25 +00:00Commented Aug 15, 2014 at 9:16
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$\begingroup$ kjetil b halvorsen he did... $\endgroup$Bak1139– Bak11392014-08-15 09:22:32 +00:00Commented Aug 15, 2014 at 9:22
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$\begingroup$ Hi , thanks! the question was : did the funtion tan(x) ,cot(x) has a symmetry axis , I tried to look at the function graph and see if there is a symmetry, i think there isn't one but I'm not sure how to get the full picture, and a good explanation why this is true without relying only on the function graph $\endgroup$Dima Shifrin– Dima Shifrin2014-08-15 09:42:11 +00:00Commented Aug 15, 2014 at 9:42
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Geometric approach
Let's first look at the graphs of $\color{blue} \tan \color{blue}x$ and $\color{red} \cot \color{red} x$. (Note: the two graphs face each other in opposite directions)
Both tangent and cotangent have origin symmetry, and this means they are odd.
Algebraic approach
Foreknowledge: $\sin x$ is odd ($\sin(-x)=-\sin(x)$), $\cos x$ is even ($\cos(-x)=\cos(x)$).
$\tan(-x)=\frac{\sin(-x)}{\cos(-x)}=\frac{-\sin(x)}{\cos(x)}=-\tan(x)$
$\cot(-x)=\frac{1}{\tan(-x)}=\frac{\cos(-x)}{\sin(-x)}=\frac{\cos(x)}{-\sin(x)}=-\cot(x)$
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$\begingroup$ the origin is a point but i was asking about a symmetry axis (tha same as y axes in cos(x)), how can i expain y=tan(x) hasn't symmetry axis? $\endgroup$Dima Shifrin– Dima Shifrin2014-08-16 06:17:18 +00:00Commented Aug 16, 2014 at 6:17
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$\begingroup$ Try applying the following formula: $\tan(x-a)=\tan(x+a)$, where $a$ is the vertical symmetry axis. If this is true for $x\in\{\text{domain of }\tan x \}$, then $x=a$ is the symmetry axis. $\endgroup$rae306– rae3062014-08-16 08:11:57 +00:00Commented Aug 16, 2014 at 8:11