Trigonometric Identities Question 112
Question: If $ p_{n}={{\cos }^{n}}\theta +{{\sin }^{n}}\theta , $ then $ ^{pn-p}n-2{{=}^{kp}}n-4 $ where:
Options:
A) $ k=1 $
B) $ k=-{{\sin }^{2}}\theta {{\cos }^{2}}\theta $
C) $ k={{\sin }^{2}}\theta $
D) $ k={{\cos }^{2}}\theta $
Show Answer
Answer:
Correct Answer: B
Solution:
$ p_{n}-{p_{n-2}}=({{\cos }^{n}}\theta +{{\sin }^{n}}\theta )-({{\cos }^{n-2}}\theta +{{\sin }^{n-2}}\theta ) $ $ ={{\cos }^{n-2}}\theta ({{\cos }^{2}}\theta -1)+{{\sin }^{n-2}}\theta ({{\sin }^{2}}\theta -1) $ $ =-{{\sin }^{2}}\theta {{\cos }^{n-2}}\theta -{{\cos }^{2}}\theta {{\sin }^{n-2}}\theta $ $ =-{{\sin }^{2}}\theta {{\cos }^{2}}\theta ({{\cos }^{n-4}}\theta +{{\sin }^{n-4}}\theta ) $ $ ={{\sin }^{2}}\theta {{\cos }^{2}}\theta {p_{n-4}}=k{p_{n-4}} $
$ \Rightarrow k=-{{\sin }^{2}}\theta {{\cos }^{2}}\theta $
BETA
