यदि `A=[(cos theta, sin theta),(-sin theta,cos theta)]` तो दिखाइए कि `A^(2)=[(cos 2 theta, sin 2 theta),(-sin 2 theta, cos 2 theta)]`|

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)]

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

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

cos theta+sin theta=cos2 theta+sin2 theta

((sin theta+cos theta)^(2)-1)/(sin theta*cos theta)

(sin theta-cos theta)^(2)*(sin theta+cos theta)^(2)=

sin theta+sin2 theta+sin3 theta=1+cos theta+cos2 theta

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

Evaluate the following determinants: (b) |(cos theta, -sin theta),(sin theta, cos theta)| = cos theta (cos theta) - sin theta(-sin theta) = cos^(2) theta + sin^(2) theta = 1

(cos theta + sin theta) ^ (2) + (cos theta-sin theta) ^ (2) =