Answer
See work step by step
Work Step by Step
Get the product:
$\left[\begin{array}{ccc}1 & -\tan \varphi / 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] \cdot\left[\begin{array}{ccc}1 & 0 & 0 \\ \sin \varphi & 1 & 0 \\ 0 & 0 & 1\end{array}\right] \cdot\left[\begin{array}{ccc}1 & -\tan \varphi / 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=$
$=\left[\begin{array}{ccc}1-\tan (\varphi / 2) \sin \varphi & -\tan \varphi / 2 & 0 \\ \sin \varphi & 1 & 0 \\ 0 & 0 & 1\end{array}\right] \cdot\left[\begin{array}{ccc}1 & -\tan \varphi / 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=$
Use identity $\tan \frac{\varphi}{2}=\frac{1-\cos \varphi}{\sin \varphi}=\frac{\sin \varphi}{1+\cos \varphi}$ to simplify expression $1-$
$\tan (\varphi / 2) \sin \varphi$
$1-\tan (\varphi / 2) \sin \varphi=1-\frac{1-\cos \varphi}{\sin \varphi} \sin \varphi=\cos \varphi$
Use this result in matrix we get:
$=\left[\begin{array}{ccc}\cos \varphi & -\tan \varphi / 2 & 0 \\ \sin \varphi & 1 & 0 \\ 0 & 0 & 1\end{array}\right] \cdot\left[\begin{array}{ccc}1 & -\tan \varphi / 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=$
$=\left[\begin{array}{ccc}\cos \varphi & \cos \varphi(-\tan \varphi / 2)-\tan \varphi / 2 & 0 \\ \sin \varphi & -\sin \varphi \tan \varphi / 2+1 & 0 \\ 0 & 0 & 1\end{array}\right]$
2
Using the same identity, we can get that:
$\cos \varphi(-\tan \varphi / 2)-\tan \varphi / 2=-(\cos \varphi+1) \tan \varphi / 2=-(\cos \varphi+1) \frac{\sin \varphi}{1+\cos \varphi}=$
$-\sin \varphi$
So, final matrix becomes:
$\left[\begin{array}{ccc}\cos \varphi & -\sin \varphi & 0 \\ \sin \varphi & \cos \varphi & 0 \\ 0 & 0 & 1\end{array}\right]$
which is the transformation matrix in homogeneous coordinates for a rotation in $\mathbb{R}^{2}$
