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Question
Show that f(x) f(y) = f(x + y), where f(x) = `[(cosx, -sinx, 0),(sinx, cosx, 0),(0, 0, 1)]`
Solution
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)]`
f(x) f(y) = `[(cosx, -sinx, 0),(sinx, cosx, 0),(0, 0, 1)] [(cosy, -siny, 0),(siny, cosy, 0),(0, 0, 1)]`
= `[(cosx cosy - sinx siny + 0, -cosx siny - sinx cosy + 0, 0 + 0 + 0),(sinx cosy + cosx siny + 0, -sinx siny + cosx cosy + 0, 0 + 0 + 0),(0 + 0 + 0, 0 + 0 + 0, 0 + 0 + 1)]`
= `[(cosx cosy - sinx siny, -(sinx cosy + cosx siny), 0),(sinx cosy + siny, (cosx cosy - sinx siny), 0),(0, 0, 1)]`
f(x) f(y) = `[(cos(x + y), -sin(x + y), 0),(sin(x + y), cos(x + y), 0),(0, 0, 1)]`
f(x) f(y) = f(x + y)
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