Question

If F(x) = `[(cosx, -sinx,0), (sinx, cosx, 0),(0,0,1)]`,  show that F(x)F(y) = F(x + y).

Answer 1

Here F(x) = `[(cosx, -sinx, 0),(sinx, cosx, 0),(0, 0, 1)]`

∴ F(y) = `[(cosy, -siny, 0),(siny, cosy, 0),(0, 0, 1)]`

∴ F(x + y) = `[(cos(x + y), -sin(x + y), 0), (sin(x + y), cos(x + y), 0),(0,0,1)]`

Now, L.H.S. = F(x).F(y)

= `[(cosx, -sinx, 0),(sinx, cosx, 0),(0, 0, 1)][(cosy, -siny, 0),(siny, cosy, 0),(0, 0, 1)]`

= `[(cosxcosy - sinxsiny, -cosxsiny - sinxsiny, 0),(sinxcosy + cosxsiny, -sinxsiny + cosxcosy, 0),(0, 0, 1)]`

= `[(cos (x + y), -sin (x + y), 0), (sin (x + y), cos (x + y), 0), (0, 0, 1)]`

= F(x + y) = R.H.S.