Question

If $A_{\alpha }=\left[\begin{array}{cc}cosn\alpha & sinn\alpha \\ -sinn\alpha & cosn\alpha \end{array}\right]$, then prove the following
$(A_{\alpha }{)}^n=\left[\begin{array}{cc}cosn\alpha & sinn\alpha \\ -sinn\alpha & cosn\alpha \end{array}\right]$

Answer 1

$(A_{\alpha }{)}^2=A_{\alpha }{A}_{\alpha }\left[\begin{array}{cc}cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{array}\right]$
$\left[\begin{array}{cc}co{s}^2\alpha -si{n}^2\alpha & 2sin\alpha cos\alpha \\ -2sin\alpha cos\alpha & -si{n}^2\alpha +co{s}^2\alpha \end{array}\right]$
$\left[\begin{array}{cc}cos2\alpha & sin2\alpha \\ -sin2\alpha & cos2\alpha \end{array}\right]$
Similarly, $(A_{\alpha }{)}^3=(A_{\alpha }{)}^{2}A\alpha$
$\left[\begin{array}{cc}cos2\alpha & sin2\alpha \\ -sin2\alpha & cos2\alpha \end{array}\right]\left[\begin{array}{cc}cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{array}\right]$
$\left[\begin{array}{cc}cos(2\alpha +\alpha )& sin(2\alpha +\alpha )\\ -sin(2\alpha +\alpha )& cos(2\alpha +\alpha )\end{array}\right]$
$\left[\begin{array}{cc}cos3\alpha & sin3\alpha \\ -sin3\alpha & cos3\alpha \end{array}\right]$
In the light above let us assume that
$(A_{\alpha }{)}^n=\left[\begin{array}{cc}cosn\alpha & sinn\alpha \\ -sinn\alpha & cosn\alpha \end{array}\right]$
$(A_{\alpha }{)}^{n+1}=(A_{\alpha }{)}^{n}A\alpha$
$=\left[\begin{array}{cc}cosn\alpha & sinn\alpha \\ -sinn\alpha & cosn\alpha \end{array}\right]$ $\left[\begin{array}{cc}cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{array}\right]$
= $\left[\begin{array}{cc}cos(n+1)\alpha & sin(n+1)\alpha \\ -sin(n+1)\alpha & cos(n+1)\alpha \end{array}\right]$
Thus we observe that our assumption for $(A_{\alpha }{)}^n$ is true for n = 2,3..and hence it is true universally .