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Home > Mathematics > Trigonometry > Tangent is an odd function tan(-x)=-tan x
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We prove here that the tangent function tan (-x) = - tan x is odd using the unit circle.
We start with the following configuration:
– unit circle $\mathcal{C}(O,R=1)$
– definition of the angle $x$
– definition of the angle $-x$
Now consider the triangles: $(OA_xA)$ and $(OA’_xA’)$.
Take the definition of the tangent of the angles $x$ and $-x$.
In the triangle $(OA_xA)$ :
$$ \tan x=\frac{\textrm{opposite}}{\textrm{adjacent}}=\frac{|OA_y|}{|OA_x|} $$
In the triangle $(OA’_xA’)$:
$$ \tan (-x)=\frac{\textrm{opposite}}{\textrm{adjacent}}=\frac{|OA’_y|}{|OA_x|} $$
By construction: $|OA_y|= -|OA’_y|$, then we have
$$ \forall x\in \mathbb{R},x \neq k\pi + \frac{\pi}{2} , k \in \mathbb{Z}:\quad \tan (-x)=\frac{|OA’_y|}{|OA_x|}=-\frac{|OA_y|}{|OA_x|}=-\tan x $$
The definition of an odd function is that $f(-x) = -f(x)$ for all $x$ in the domain of the function. To prove that the tangent function is odd, we can use this definition and show that $\tan(-x) = -\tan x$ for all $x \neq k\pi + \frac{\pi}{2} , k \in \mathbb{Z}$
Using the trigonometric identity for tangent, we can write:
$$\tan(-x) = \frac{\sin(-x)}{\cos(-x)}$$
Using the fact that sine is odd $\sin(-x) = -\sin x$ and cosine is even $\cos(-x) = \cos x$, we can simplify the above equation to:
$$\tan(-x) = -\left(\frac{\sin x }{\cos x}\right) = -\tan x$$
which means that the tangent function is an odd function.