Cosine is even function cos(-x)=cos x

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 $(OA’_xA’)$.

Proof that cosine is an even function cos(-x) = cos (x)

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

$$ \forall x\in \mathbb{R},\quad: \cos (-x)=\cos x $$