> 
> Show that every rotation in $\mathbb{R^3}$ can be written as the
>  product of two rotations of order 2.
> 
> 
> 

Here's my attempt at a solution:

We know that any rotation in $\mathbb{R^3}$ can be represented as the product of two reflections. So we write our rotation as $R\_1R\_2$ where the $R\_j$ are reflections in $\mathbb{R^3}$.

We would like to show that $R\_1R\_2$=$(R\_aR\_b)(R\_cR\_d)$=$R\_aR\_bR\_cR\_d$ where $R\_aR\_b$ and $R\_cR\_d$ are rotations of order 2 in $\mathbb{R^3}$.
I think I have shown that a rotation $R\_1R\_2$, which is the product of reflections in the planes $\Pi\_1$ and $\Pi\_2$ respectively, has order two if and only if $\Pi\_1$ and $\Pi\_2$ are perpendicular. Next I thought it sufficient to show that $R\_1$=$R\_aR\_b$ (and similarly for $R\_2$ with $R\_c$ and $R\_d$) and as $R\_1$ and $R\_2$ are just any reflections, I now attempt to show that any reflection can be written as the product of two reflections in perpendicular planes.

The form of a reflection in the plane $x.n=d$ in $\mathbb{R^3}$ is given as $R(x)=x+2(d-x.n)n$.

Suppose that we have $R\_a$ and $R\_b$ as reflections in the perpendicular planes (that is, the planes have perpendicular normals), $x.n\_a=d\_a$ and $x.n\_b=d\_b$ respectively. As $n\_a.n\_b=0$, we can show that $R\_aR\_b(x)= x+2(d\_1+d\_2)-(x.(n\_1+n\_2))(n\_1+n\_2)$ which is of the required form. So that for any reflection $R$ in $x.n=d$, we can set $n\_1,n\_2,d\_1,d\_2$ such that $d\_1+d\_2=d$ and $n\_1+n\_2=n$.

First off, is this correct, have I shown what I was required to show? I was also wondering if there may be a nicer, perhaps geometric way of going about the question.

Apologies if my attempt at a solution is difficult to follow, I have next to zero experience in writing formal solutions.