Question

If $F\left(x\right)=\left[\begin{array}{ccc}cosx& -sinx& 0\\ sinx& cosx& 0\\ 0& 0& 1\end{array}\right]$, then show that F (x)F(y)=F(x+y).

Answer 1

$LHS=F\left(x\right)F\left(y\right)=\left[\begin{array}{ccc}cosx& -sinx& 0\\ sinx& cosx& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}cosy& -siny& 0\\ siny& cosy& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}cosxcosy-sinxsiny& -sinycosx-sinxcosy& 0\\ sinxcosy+cosxsiny& -sinxsiny+cosxcosy& 0\\ 0& 0& 1\end{array}\right]$

$\left[\begin{array}{c}\because cos(A+B)=cosAcosB-sinAsinB\\ sin(A+B)=sinAcosB+sinBcosA\end{array}\right]=\left[\begin{array}{ccc}cos(x+y)& -sin(x+y)& 0\\ sin(x+y)& cos(x+y)& 0\\ 0& 0& 1\end{array}\right]$
Now, replacing x by (x+y)in F(x) F(y)
$\therefore F(x+y)=\left[\begin{array}{ccc}cos(x+y)& -sin(x+y)& 0\\ sin(x+y)& cos(x+y)& 0\\ 0& 0& 1\end{array}\right]F\left(x\right)F\left(y\right)=F(x+y)=RHS$. Hence, proved.