Question

If $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$ then show that for all the positive integers $n$ $A^n=\left[\begin{array}{cc}\cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{array}\right]$

Answer 1

$A=\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)$

$A^2=\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)$

$=\left(\begin{array}{cc}{\cos }^2\theta -{\sin }^2\theta & \sin \theta \cos \theta +\sin \theta \cos \theta \\ -\cos \theta \sin \theta -\cos \theta \sin \theta & {\cos }^2\theta -{\sin }^2\theta \end{array}\right)$

$=\left(\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{array}\right)$

The result is true for $n=1$ and $2$

Let us assume the result is true for $n=k$

i.e., $A^k=\left(\begin{array}{cc}\cos k\theta & \sin k\theta \\ -\sin k\theta & \cos k\theta \end{array}\right)$

We have to prove the result for $n=k+1$

$A^{k+1}=A^k\cdot A$

$=\left(\begin{array}{cc}\cos k\theta & \sin k\theta \\ -\sin k\theta & \cos k\theta \end{array}\right)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)$

$=\left(\begin{array}{cc}\cos k\theta \cos \theta -\sin k\theta \sin \theta & \cos k\theta \sin \theta +\sin k\theta \cos \theta \\ -(\sin k\theta \cos \theta +\cos k\theta \sin \theta )& \cos k\theta \cos \theta -\sin k\theta \sin \theta \end{array}\right)$

$=\left(\begin{array}{cc}\cos (k+1)\theta & \sin (k+1)\theta \\ -\sin (k+1)\theta & \cos (k+1)\theta \end{array}\right)$

The result is true for $n=k+1$ also. Hence by the principle of Mathematical Induction, we can understand that the result is true for all $n\in N$.