Trigonometric Identities Question 355
Question: $ {{\cos }^{2}}48{}^\circ -{{\sin }^{2}}12{}^\circ = $
[MNR 1977]
Options:
A) $ \frac{\sqrt{5}-1}{4} $
B) $ \frac{\sqrt{5}+1}{8} $
C) $ \frac{\sqrt{3}-1}{4} $
D) $ \frac{\sqrt{3}+1}{2\sqrt{2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ {{\cos }^{2}}A-{{\sin }^{2}}B=\cos ,(A+B),.,\cos ,(A-B) $
$ \therefore {{\cos }^{2}}48^{o}-{{\sin }^{2}}12^{o}=\cos 60^{o},.,\cos 36^{o} $ $ =\frac{1}{2},( \frac{\sqrt{5}+1}{4} )=\frac{\sqrt{5}+1}{8}. $
BETA
