Trace identity for $2 \times 2$ reflections Let $A, B, C \in \mathrm{GL}(2,\mathbb{C})$ be reflections (i.e., their eigenvalues are $\pm 1$). Please show that
$$ \DeclareMathOperator\Tr{Tr}\{\Tr(AB)\}^2+\{\Tr(BC)\}^2 + \{\Tr(CA)\}^2 - \{\Tr(AB)\}\{\Tr(BC)\}\{\Tr(CA)\}\\ = 2 + \Tr\left((ABC)^2\right) $$
holds.
 A: Here is a simple proof that can be done without Mathematica (and without trigonometry).
A general $2\times2$ reflection matrix is of the form
$$M(a,t):=\begin{bmatrix}a&t\\(1-a^2)/t&-a\end{bmatrix}$$
with some complex $a,t$ (or the limit, $\pm\begin{bmatrix}1&0\\z&-1\end{bmatrix}$ for any $z\in\mathbb C$, of $M(\pm\sqrt{1-tz},t)$ as $0\ne t\to0$). This follows because a reflection matrix has trace $0$ and determinant $-1$.
By diagonalization, without loss of generality (wlog)
$$A=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$$
and, by continuity, we can also write
$$B=\begin{bmatrix}b&u\\(1-b^2)/u&-b\end{bmatrix}\quad\text{and}\quad 
C=\begin{bmatrix}c&v\\(1-c^2)/v&-c\end{bmatrix}$$
with some complex $b,u,c,v$.
Substituting these expressions for $A,B,C$ in the identity in question and simplifying, we verify the identity.

This argument should work for $2\times2$ reflection matrices $A,B,C$ over any field -- except that, instead of the limit transition, one would have to consider separately the cases when at least one of the matrices $B,C$ is lower triangular; but those cases should be even easier to check than the "main" case, when $B$ and $C$ are not lower triangular.
A: $\DeclareMathOperator\Tr{Tr}$For real matrices $A$, $B$, $C$ the proof can be done "by hand": parameterize an orthogonal reflection matrix as
$$M_n=\begin{pmatrix}\cos\theta_n&\sin\theta_n\\
\sin\theta_n&-\cos\theta_n\end{pmatrix},$$
with $A=M_1$, $B=M_2$, $C=M_3$. Note that $(M_1M_2M_3)^2=I$. Then one has
$$\Tr M_iM_j=2\cos(\theta_i-\theta_j),
\quad\Tr((M_1M_2M_3)^2)=2,$$
and the left-hand-side and right-hand-side of the equation in the OP both evaluate to 4, in view of the identity
$$\cos ^2(\theta_1-\theta_2)+\cos ^2(\theta_2-\theta_3)+\cos ^2(\theta_3-\theta_1)=1+2 \cos (\theta_1-\theta_2) \cos (\theta_2-\theta_3) \cos (\theta_3-\theta_1).$$

For complex matrices the parameterization is
$$M_n=\begin{pmatrix}\cos\theta_n&e^{i\phi_n}\sin\theta_n\\
e^{-i\phi_n}\sin\theta_n&-\cos\theta_n\end{pmatrix}.$$
Then one has
$$\Tr M_iM_j=2\cos\theta_i\cos\theta_j+2\cos(\phi_i-\phi_j)\sin\theta_i\sin\theta_j.$$
The trace of $(M_1M_2M_3)^2$ is a lengthy expression, which according to Mathematica is equivalent to
$$\Tr(M_1M_2M_3)^2=[\Tr M_1M_2]^2+[\Tr M_3M_1]^2+[\Tr M_2M_3]^2-[\Tr M_1M_2] [\Tr M_3M_1] [\Tr M_2M_3]-2.$$
