Opposite real number identities(why $\cos(-x)=\cos x$) 
Ive been studying about opposite real number identities and Ive been stuck on this question on why $\cos(-x)=\cos x$.
Okay so if we consider that the given circle is a unit circle and triangle $pom$ and triangle $qom$ are congruent then how $\cos(-x)=\cos x$?
According to me when we will do base/hypotenuse for triangle qom then it will come $om/oq$ which should give $-\cos x$ as $oq$ is negative, right? 
Please help me through this.

A non-geometry approach would be to consider the series definition for cosine. With this, for all $x\in\mathbb R$ the series $\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}$ converges and
$$\cos(x)=\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}.$$
For every $x\in\mathbb R$, you can now trivially see that
$$\cos(-x)=\sum_{n=0}^\infty\frac{(-1)^n(-x)^{2n}}{(2n)!}=\sum_{n=0}^\infty\frac{(-1)^n(-1)^{2n}x^{2n}}{(2n)!}=\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}=\cos(x)$$
