e ,
P(k + 1) : Ak+1 =
cos (
1)
sin (
1)
sin(
1)
cos (
1 )
k
k
k
k
+
θ
+
θ




−
+
θ
+
θ


Now
Ak+1 = Ak A
Since P(k) is true, we have
Ak+1 =
cos
sin
sin
cos
k
k
k
k
θ
θ




−
θ
θ

   
cos
sin
sin
cos
θ
θ




−
θ
θ


= 
cos
cos
sin
sin
cos
sin
sin
cos
sin
cos
cos
sin
sin
sin
cos
cos
k
k
k
k
k
k
k
k
θ
θ −
θ
θ
θ
θ +
θ
θ




−
θ
θ −
θ
θ
−
θ
θ +
θ
θ


(by matrix multiplication)
= 
cos (
1)
sin (
1)
sin(
1)
cos (
1 )
k
k
k
k
+
θ
+
θ




−
+
θ
+
θ


Thus, P(k + 1) is true whenever P(k) is true Hence, P(n) is true for all n ≥ 1 (by the principle of mathematical induction)