Question

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

Answer 1

Given: $F\left(x\right)=\left[\begin{array}{ccc}\cos x& -\sin x& 0\\ \sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]$
To show: $F\left(x\right)F\left(y\right)=F(x+y)$
Replacing $x$ by $y$ in $F\left(x\right)$
$F\left(y\right)=\left[\begin{array}{ccc}\cos y& -\sin y& 0\\ \sin y& \cos y& 0\\ 0& 0& 1\end{array}\right]$

Solving $L.H.S.$
$F\left(x\right)F\left(y\right)$
$=\left[\begin{array}{ccc}\cos x& -\sin x& 0\\ \sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}\cos y& -\sin y& 0\\ \sin y& \cos y& 0\\ 0& 0& 1\end{array}\right]$

$F\left(x\right)F\left(y\right)=\left[\begin{array}{ccc}(\cos x\cos y-\sin x\sin y)& (-\cos x\sin y-\sin x\cos y)& 0\\ (\sin x\cos y+\cos x\sin y)& (-\sin x\sin y+\cos x\cos y)& 0\\ 0& 0& 1\end{array}\right]$
We know that
$\left[\begin{array}{c}\cos x\cos y-\sin x\sin y=\cos (x+y)\\ \sin x\cos y+\cos x\sin y=\sin (x+y)\end{array}\right]$
$F\left(x\right)F\left(y\right)=\left[\begin{array}{ccc}\cos (x+y)& -\sin (x+y)& 0\\ \sin (x+y)& \cos (x+y)& 0\\ 0& 0& 1\end{array}\right]$

Solving $R.H.S.$
$F(x+y)$
Replacing $x$ by $(x+y)$ in $F\left(x\right)$

$F(x+y)=\left[\begin{array}{ccc}\cos (x+y)& -\sin (x+y)& 0\\ \sin (x+y)& \cos (x+y)& 0\\ 0& 0& 1\end{array}\right]=L.H.S.$
Hence proved