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Question
If A = `[(costheta, sintheta),(-sintheta, costheta)]`, then show that A2 = `[(cos2theta, sin2theta),(-sin2theta, cos2theta)]`
Solution
We have, A = `[(costheta, sintheta),(-sintheta, costheta)]`
∴ A2 = A · A
= `[(costheta, sintheta),(-sintheta, costheta)] [(costheta, sintheta),(-sinteta, costheta)]`
= `[(cos^2theta - sin^2theta, costhetasintheta + sinthetacostheta),(-sinthetacostheta - costhetasintheta, -sin^2theta + cos^2theta)]`
= `[(cos2theta, 2sinthetacostheta),(2sinthetacostheta, cos2theta)]`
= `[(cos2theta, sin 2theta),(-sin2theta, cos 2theta)]`
Hence proved.
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