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

If cos theta + sin theta =sqrt2 cos theta , then the value of (cos theta -sin theta)/(sin theta) is

Verify that [(cos theta, sin theta),(-sin theta, cos theta)] and [(cos theta, - sin theta),(sin theta, cos theta)] are inverse of each other.

If A = [(cos theta, sin theta),(-sin theta, cos theta)] then show that for all the positive integers n, A^(n) = [(cos n theta,sin n theta),(-sin n theta,cos n theta)] .

If A=[(cos theta, sin theta),(-sin theta, cos theta)] , then prove that A^n=[[cosntheta,sin ntheta],[-sin ntheta,cos ntheta]], n in N

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)]

Simplify cos theta[(cos theta,sin theta),(-sin theta,cos theta)]+sin theta[(sin theta,-cos theta),(cos theta,sin theta)]

If A = [(cos theta, sin theta),(-sin theta, cos theta)] and B = [(sin theta, - cos theta), (cos theta, sin theta)] , evaluate A cos theta + B sin theta .