As the output to some calculation I get an orthogonal 2-D matrix, which I know must represent some 2-D rotation (and possibly a reflection). It looks like
\begin{bmatrix} f & g \newline 1 & 1 \end{bmatrix}
Where $f$ and $g$ are complicated trig functions. In case it is important this comes out of the singular-value decomposition of another matrix. If I am to just naïvely compare this to a general rotation matrix
\begin{bmatrix} \cos\theta & -\sin\theta \newline \sin\theta & \cos\theta \end{bmatrix}
I get $\cos\theta=\sin\theta$ so $\theta=\pi/4$ (the matrix has some missing overall constant). The problem is that I have physical reasons for thinking this can't be right.
Is there a way that this might represent a matrix for a different rotation? (Possibly also involving a reflection?)
As Ted Shifrin pointed out, the matrix explicitly given cannot be orthogonal, because the second row is not a unit vactor. The only scalars $\ c\ $ for which the matrix $$ c\begin{bmatrix} f&g\\1&1 \end{bmatrix}\tag{1}\label{e1} $$ can be orthogonal are $\ c=\pm2^{-\frac{1}{2}}\ $, and then you would need $\ fg=-1\ $ (which you have acknowledged to hold) and $\ f^2+g^2=2\ $. Combining these two restrictions on $\ f\ $ and $\ g\ $ gives you $$ f^2+\frac{1}{f^2}=2\ , $$ or, equivalently, \begin{align} 0&=f^4-2f^2+1\\ &=\big(f^2-1\big)^2\ , \end{align} so $\ f^2=1\ $. Thus, the only orthogonal matrices of the form \ref{e1} are $$ \frac{1}{\sqrt{2}}\begin{bmatrix} 1&-1\\1&1 \end{bmatrix},\ \frac{1}{\sqrt{2}}\begin{bmatrix} -1&1\\1&1\end{bmatrix},\ \frac{1}{\sqrt{2}}\begin{bmatrix} -1&1\\-1&-1\end{bmatrix} $$ and $$ \frac{1}{\sqrt{2}}\begin{bmatrix} 1&-1\\-1&-1\end{bmatrix}\ . $$