Trigonometric-Identities Question 398
Question: The value of $ {{\cos }^{2}}\frac{\pi }{12}+{{\cos }^{2}}\frac{\pi }{4}+{{\cos }^{2}}\frac{5\pi }{12} $ is
[Karnataka CET 2002]
Options:
A) $ \frac{3}{2} $
B) $ \frac{2}{3} $
C) $ \frac{3+\sqrt{3}}{2} $
D) $ \frac{2}{3+\sqrt{3}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ {{\cos }^{2}}\frac{\pi }{12}+{{\cos }^{2}}\frac{\pi }{4}+{{\cos }^{2}}\frac{5\pi }{12} $ $ =1-{{\sin }^{2}}( \frac{\pi }{12} )+{{( \frac{1}{\sqrt{2}} )}^{2}}+{{\cos }^{2}}( \frac{5\pi }{12} ) $ $ =1+\frac{1}{2}+( {{\cos }^{2}}\frac{5\pi }{12}-{{\sin }^{2}}\frac{\pi }{12} ) $ $ =\frac{3}{2}+\cos ( \frac{5\pi }{12}+\frac{\pi }{12} )\cos ( \frac{5\pi }{12}-\frac{\pi }{12} )=\frac{3}{2}+\cos \frac{\pi }{2}\cos \frac{\pi }{3} $ $ =\frac{3}{2}+0.\frac{1}{2}=\frac{3}{2} $ .
BETA
