If `A=[costhetasintheta-sinthetacostheta]`, then prove that `A^n=[cosnthetasinntheta-sinnthetacosntheta], n in N`.

Explanation

Let `P(n) equiv A^(n)=[(cos n theta,sin n theta),(-sin n theta,cos n theta)]`
For `n=1, A^(1)=[(cos theta ,sin theta),(-sin theta,cos theta)]`
For `n=2, A^(2)=AxxA`
`=[(cos theta,sin theta),(-sin theta,cos theta)][(cos theta,sin theta),(-sin theta,cos theta)]`
`=[(cos^(2) theta-sin^(2) theta,cos theta sin theta+sin theta cos theta),(-sin theta cos theta-cos theta sin theta,-sin^(2) theta+cos^(2) theta)]`
`=[(cos 2 theta,2 sin theta cos theta),(-2 sin theta cos theta,cos 2 theta)]`
`=[(cos 2 theta,sin 2 theta),(-si 2 theta,cos 2 theta)]`
Thus, P(n) is true `n=1` and `n=2`.
Let P(n) be true for some `n=k`.
`:. A^(k)=[(cos k theta,sin k theta),(-sin k theta,cos k theta)]`
`:. A^(k+1)=A^(k)A`
`=[(cos k theta,sin k theta),(-sin k theta,cos k theta)][(cos theta,sin theta),(- sin theta,cos theta)]`
`=[(cos k theta cos theta - sin k theta sin theta,cos k theta sin theta +sin k theta cos theta),(-sin k theta cos theta - cos k theta sin theta,-sin k theta sin theta + cos k theta cos theta)]`
`=[(cos(k+1)theta,sin (k+1) theta),(-sin (k+1) theta,cos (k+1) theta)]`
Thus, P(n) is true for `n=k+1` also.
Hence, P(n) is true for all `n in N`.