Question

If A = `[(costheta, sintheta),(-sintheta, costheta)]`, then show that A² = `[(cos2theta, sin2theta),(-sin2theta, cos2theta)]`

Answer 1

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.