Trigonometric Identities Question 184
Question: If $ \sin x+{{\sin }^{2}}x=1 $ , then the value of $ {{\cos }^{12}}x+3{{\cos }^{10}}x+3{{\cos }^{8}}x+{{\cos }^{6}}x-2 $ is equal to
[Pb. CET 2002]
Options:
A) 0
B) 1
C) - 1
D) 2
Show Answer
Answer:
Correct Answer: C
Solution:
We have, $ \sin x+{{\sin }^{2}}x=1 $ or $ \sin x=1-{{\sin }^{2}}x $ or $ \sin x={{\cos }^{2}}x $ \ $ {{\cos }^{12}}x+3{{\cos }^{10}}x+3{{\cos }^{8}}x+{{\cos }^{6}}x-2 $ $ ={{\sin }^{6}}x+3{{\sin }^{5}}x+3{{\sin }^{4}}x+{{\sin }^{3}}x-2 $ $ ={{({{\sin }^{2}}x)}^{3}}+3{{({{\sin }^{2}}x)}^{2}}\sin x $ $ +3({{\sin }^{2}}x){{(\sin x)}^{2}}+{{(\sin x)}^{3}}-2 $ $ ={{({{\sin }^{2}}x+\sin x)}^{3}}-2 $ $ ={{(1)}^{3}}-2 $ $ [\because \sin x+{{\sin }^{2}}x=1(given)] $ = - 1.
BETA
