Question
If $${p_n} = {\cos ^n}\theta + {\sin ^n}\theta ,$$ then $${p_n} - {p_{n - 2}} = k{p_{n - 4}},$$ where :
A.
$$k = 1$$
B.
$$k = - {\sin ^2}\theta \,{\cos ^2}\theta $$
C.
$$k = {\sin ^2}\theta $$
D.
$$k = {\cos ^2}\theta $$
Answer :
$$k = - {\sin ^2}\theta \,{\cos ^2}\theta $$
Solution :
$$\eqalign{
& {p_n} - {p_{n - 2}} = \left( {{{\cos }^n}\theta + {{\sin }^n}\theta } \right) - \left( {{{\cos }^{n - 2}}\theta + {{\sin }^{n - 2}}\theta } \right) \cr
& = {\cos ^{n - 2}}\theta \left( {{{\cos }^2}\theta - 1} \right) + {\sin ^{n - 2}}\theta \left( {{{\sin }^2}\theta - 1} \right) \cr
& = - {\sin ^2}\theta \,{\cos ^{n - 2}}\theta - {\cos ^2}\theta \,{\sin ^{n - 2}}\theta \cr
& = - {\sin ^2}\theta \,{\cos ^2}\theta \left( {{{\cos }^{n - 4}}\theta + {{\sin }^{n - 4}}\theta } \right) \cr
& = - {\sin ^2}\theta \,{\cos ^2}\theta {p_{n - 4}} = k{p_{n - 4}} \cr
& \Rightarrow k = - {\sin ^2}\theta \,{\cos ^2}\theta \cr} $$