If `A=[[costheta,isintheta],[isintheta,costheta]],` then prove by principal of mathematical induction that `A^n=[[cosntheta, i sinn theta],[isin n theta, cos n theta]]` for all n`in` `NN`.

Explanation

Here, it is given that,

`A = [[cos theta, isintheta],[isintheta,cos theta]]`.

We have to prove,

`A^n = [[cos ntheta, isin ntheta],[isin ntheta,cos ntheta]]`.

Now, for `n = 1`.

`A^1 = [[cos 1theta, isin 1theta],[isin 1theta,cos 1theta]]`.

`=>A = [[cos theta, isintheta],[isintheta,cos theta]]`, which is true.

So, given equation is true for `n = 1`.

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