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

Inverse of the matrix [{:(cos 2 theta, -sin 2 theta),(sin 2 theta, cos 2 theta):}] is

Inverse of the matrix [(cos 2 theta,-sin 2theta),(sin 2 theta, cos 2theta)] is

cos theta+sin theta=cos2 theta+sin2 theta

If f (theta) = [[cos^(2) theta , cos theta sin theta,-sin theta],[cos theta sin theta , sin^(2) theta , cos theta ],[sin theta ,-cos theta , 0]] "then " f (pi/7) is

If A(theta) = [(cos theta,sin theta),(-sin theta,cos theta)] then prove that A^(2)(theta) = A(2 theta)

if A=[{:(costheta,sin theta ),(-sin theta,costheta):}] , then show that : A^(2)=[{:(cos2theta,sin2theta),(-sin2theta,cos2theta):}]