Question

If $F\left(x\right)=[\cos x-\sin x0$,

$\sin x\cos x0$

$0,0,1$]

show that $F\left(x\right)F\left(y\right)=F(x+y)$

Answer 1

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

So, $F\left(x\right).F\left(y\right)=F(x+y)$

Hence Proved