If `A=[[cos 2 theta, sin 2 theta],[-sin2theta, cos2theta]]`, find `A^2`

If A=[[cos2 theta,sin2 theta],[-sin2 theta,cos2 theta]] then find A^2

If A=[cos2 theta sin2 theta-sin2 theta cos2 theta], find A^(2)

Inverse of the matrix A=[[cos2theta, -sin2theta], [sin2theta, cos2theta]] is

sin2theta - cos2theta =1

If A=[[costheta,sintheta],[-sintheta,costheta]] then prove that A^2=[[cos2theta,sin2theta],[-sin2theta,cos2theta]]

The inverse of the matrix [(cos2theta, -sin2theta),(sin2theta, cos2theta)] is -

Let A = [ (cos theta, sin theta),(- sin theta, cos theta)] , then show that A ^(2) = [(cos 2 theta, sin 2 theta),( - sin 2 theta, cos 2 theta)]

cos theta+sin theta=cos2 theta+sin2 theta

(1-cos2theta)/(sin2theta)=