I have read that, given any root system $\Phi$ of a real Euclidean vector space $V$, every (simple) reflection $s_{\alpha} \colon V \to V$ has $$\det s_{\alpha} = -1.$$
I understand why $\det s_{\alpha}$ must be $\pm 1$: $s_{\alpha}$ satisfies $s_{\alpha}^2 = 1$. But why must its determinant be $-1$?
It is easily seen that $s_\alpha$ is diagonalisable with eigenvectors $\pm 1$, where the eigenspace for the eigenvalue $-1$ is precisely 1-dimensional; from this, the claim follows.
Namely, I assume the definition of $s_\alpha$ is $s_\alpha(v) =v - \check{\alpha}(v)\cdot \alpha$ where $\check{\alpha} \in V^*$ with $\check{\alpha}(\alpha) =2$. Then
$s_\alpha(\alpha) = -\alpha$ and
$s_\alpha(v) = v$ for all $v \in \ker(\check{\alpha})$, which is a $(\dim V-1)$-dimensional subspace of $V$ (clear?).
Now if $(v_2, ..., v_n)$ is any basis of $\ker(\check{\alpha})$, then $(\alpha, v_2, ..., v_n)$ is a basis of $V$ (clear?) and with respect to that basis, $s_\alpha$ corresponds to the diagonal matrix with diagonal entries $(-1, 1, ...., 1)$.
If your definition of $s_\alpha$ is different, the argument w.r.t. that definition should be just as easy.