Question

Given $F\left(x\right)=\left[\begin{array}{ccc}\cos x& -\sin x& 0\\ \sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]$ If $x\in R$ Then prove $y$, $F(x+y)=F\left(x\right)F\left(y\right)$

Answer 1

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

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

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

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

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