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Home > Mathematics > Trigonometry > Cosine is even function cos(-x)=cos x
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We prove here that the cosine function cos(-x)=cos x is even 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 cosines of the angles $x$ and $-x$.
In the triangle $(OA_xA)$ :
$$ \cos x=\frac{\textrm{adjacent}}{\textrm{hypotenuse}}=\frac{|OA_x|}{R}=\frac{|OA_x|}{1}=|OA_x| $$
In the triangle $(OA’_xA’)$:
$$ \cos (-x)=\frac{\textrm{adjacent}}{\textrm{hypotenuse}}=\frac{|OA’_x|}{R}=\frac{|OA’_x|}{1}=|OA’_x| $$
By construction: $|OA_x|= |OA’_x|$, then we have
$$ \forall x\in \mathbb{R},\quad: \cos (-x)=\cos x $$