Question

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`.

Answer 1

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`.
Now, let given equation is true for `n = k`.
Then, `A^k = [[cos ktheta, isin ktheta],[isin ktheta,cos ktheta]]`->(1). Now, we have to prove given equation is true for `n = k+1`.
We have to prove,
`A^(k+1) = [[cos (k+1)theta, isin (k+1)theta],[isin (k+1)theta,cos (k+1)theta]]`
Now, `L.H.S. = A^(k+1) = A^k*A = [[cos ktheta, isin ktheta],[isin ktheta,cos ktheta]] [[cos theta, isintheta],[isintheta,cos theta]]`
`= [[cos kthetacostheta + i^2sin kthetasin theta, icoskthetasin theta + isinkthetacos theta],[isin ktheta cos theta+ i cosktheta sin theta,i^2sin ktheta sin theta+cos k theta cos theta]] `
`= [[cos kthetacostheta - sin kthetasin theta, i(coskthetasin theta + sinkthetacos theta)],[i(sin ktheta cos theta+ cosktheta sin theta),-sin kthetasin theta+cos k theta cos theta]] `
`=[[cos (k+1)theta, isin (k+1)theta],[isin (k+1)theta,cos (k+1)theta]] = R.H.S.`
Thus, our equation is true for `n = k+1`.
`:.A^n = [[cos ntheta, isin ntheta],[isin ntheta,cos ntheta]]`.