Trigonometric Identities Question 245
Question: Let $ x+y=3-\cos 4\theta $ and $ x-y=4sin2\theta $ then the greatest of $ xy $ is
Options:
A) $ \frac{3}{4} $
B) $ 1 $
C) $ \frac{1}{2} $
D) 2
Show Answer
Answer:
Correct Answer: B
Solution:
$ x=\frac{3-\cos 4\theta +4sin2\theta }{2} $ $ =\frac{3-(1-{{\sin }^{2}}2\theta )+4\sin 2\theta }{2}={{(1+\sin 2\theta )}^{2}} $ $ y=\frac{3-\cos 4\theta -4\sin 2\theta }{2} $ $ =\frac{3-(1-{{\sin }^{2}}2\theta )+4\sin 2\theta }{2}={{(1-\sin 2\theta )}^{2}} $
$ \therefore xy={{(1-{{\sin }^{2}}2\theta )}^{2}}=\cos {{,}^{4}}2\theta \le 1 $
BETA
