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