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Question
If f(α) = `[(cosα, -sinα, 0),(sinα, cosα, 0),(0, 0, 1)]`, prove that f(α) . f(– β) = f(α – β).
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Solution
Given, f(α) = `[(cosα, -sinα, 0),(sinα, cosα, 0),(0, 0, 1)]`
f(– β) = `[(cos(–β), –sin(–β), 0),(sin(–β), cos(–β), 0),(0, 0, 1)]`
= `[(cosβ, sinβ, 0),(-sinβ, cosβ, 0),(0, 0, 1)]`
f(α) . f(– β) = `[(cosα, -sinα, 0),(sinα, cosα, 0),(0, 0, 1)][(cosβ, sinβ, 0),(-sinβ, cosβ, 0),(0, 0, 1)]`
= `[(cosα cosβ + sinα sinβ, cosα sinβ - sinα sinβ, 0),(sinα cosβ - sinβ cosα, sinα sinβ + cosα cosβ, 0),(0, 0, 1) ]`
= `[(cos(α - β), -sin(α - β), 0),(sin(α - β), cos(α - β), 0),(0, 0, 1)]`
= f(α – β).
Hence Proved.
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