Question

If f(α) = `[(cosα, -sinα, 0),(sinα, cosα, 0),(0, 0, 1)]`, prove that f(α) . f(– β) = f(α – β).

Answer 1

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.