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Question
Answer the following question:
If Aα = `[(cosalpha, sinalpha),(-sinalpha, cosalpha)]`, show that Aα . Aβ = Aα+β
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Solution
Aα = `[(cosalpha, sinalpha),(-sinalpha, cosalpha)]`
∴ Aβ = `[(cosbeta, sinbeta),(-sinbeta, cosbeta)]`
∴ Aα.Aβ = `[(cosalpha, sinalpha),(-sinalpha, cosalpha)] [(cosbeta, sinbeta),(-sinbeta, cosbeta)]`
`= [(cosalphacosbeta - sinalphasinbeta, cosalphasinbeta + sinalpha cosbeta),(-sinalpha cosbeta - cosalpha sinbeta, -sinalphasinbeta + cosalpha cosbeta)]`
`= [(cosalpha cosbeta - sinalpha sinbeta, cosalpha sinbeta + sinalpha cosbeta),(-[sinalpha cosbeta + cosalpha sinbeta], cosalphasinbeta - sinalpha cosbeta)]`
`= [(cos(alpha + beta), sin(alpha + beta)),(-sin(alpha + beta), cos(alpha + beta))]`
∴ Aα . Aβ = Aα+β
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