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