If `A=[(costheta,isintheta),(isintheta,costheta)]` , then prove by principle of mathematical induction that `A^n=[(cosn\ theta,isinn\ theta),(isinn\ theta,cosn\ theta)]` for all `n in N` .

Explanation

Given that,

`A= [(costheta,isintheta),(isintheta,costheta)]`

We have to prove that `A^n=[(cosn\ theta,isinn\ theta),(isinn\ theta,cosn\ theta)]` for all positive integers `ndot`

It is true for `n=1`.

`implies A^2= [(costheta,isintheta),(isintheta,costheta)]xx [(costheta,isintheta),(isintheta,costheta)]`

`implies A^2= [(cos2theta,isin2theta),(isin2theta,cos2theta)]`

It is true for `n=2`.

Let it is true for `n=n-1`

...