$f: S^1\rightarrow \mathbb{R}$, then there exists a point $x$ of $S^1$ such $f(x)=f(-x)$

Question

Let $f: S^1\rightarrow \mathbb{R}$ is a continuous map. Show there exists a point $x$ of $S^1$ such $f(x)=f(-x)$.

I could not understand what $-x$ means here?

I know that, $S^1=\{x\times y: \|x\times y\|=1\}$, where $x\times y\in\mathbb{R^2}$. I know the answer from this and this links.

Answer 1

$\mathbb R^2$ is a vector space and $-x$ means opposite vector. If we use coordinates, $-(x,y) = (-x,-y)$ because $$(x,y)+(-x,-y) = (x+(-x),y+(-y)) = (0,0) = (-x+x,-y+y) = (-x,-y)+(x,y).$$

Also notice that $\|-x\| = \| x\|$ and hence, if $x\in\mathbb S^1$, then so is $-x$.

Answer 2

On $\Bbb S^1$ draw a line from $x\in \Bbb S^1$ to the origin in circle. continuing this line hit the circle on $-x$.