If `P(x)=[(cosx, sinx),(-sinx, cosx)]`, then show that `P(x).P(y)=P(x+y)=P(y).P(x)`.

Explanation

We have.
`P(x)=[{:(cos x,sinx ),(-sinx,cosx):}]`
`therefore P(y)=[{:(cos y, sin y),(-siny,cosy):}]`
Now, `P(x).P(y)=[{:(cosx,sinx),(-sin x,cosx):}][{:(cosy,siny),(-siny,cosy):}]`
`=[{:(cosx.cosy-cos x.siny,-sin x.siny+cosx.cosy),(-sinx.cosy-cosx.siny,-sinx.siny+cosx.cosy):}]`
`=[{:(cos(x+y),sin(x+y)),(-sin(x+y),cos(x+y)):}]`
`[{:(because cos(x+y)=,cosx.cosy-sinx.siny),("and" sin(x+y)=,sinx.cosy+cosx.siny):}]`
and `P(x+y)=[{:(cos(x+y)),(-sin(x+y),cos(x+y )):}]`
Also `P(y).P(x)=[{:( cosy,siny),(-siny,cosy):}][{:(cosx,sinx),(-sinx,cosx):}]`
`= [{:(cosy.cosx-siny.sinx,cosy.sinx+siny.cosx),(-siny.cosx-sin x.cos y,-sin y.sin x+cos y.cos x):}]`
`=[{:(cos(x+y),sin(x+y)),(-sin(x+y),cos(x+y)):}]`
Thus, we see from the Eqs. (i), (ii) and (iii) that
`P(x).P(y)=P(x+y)=P(y).P(x)` Hence proved.