Trigonometric Identities Question 71
Question: $ {{\cos }^{2}}( \frac{\pi }{6}+\theta )-{{\sin }^{2}}( \frac{\pi }{6}-\theta )= $
[EAMCET 2001]
Options:
A) $ \frac{1}{2}\cos 2\theta $
B) 0
C) $ -\frac{1}{2}\cos 2,\theta $
D) $ \frac{1}{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ {{\cos }^{2}}( \frac{\pi }{6}+\theta )-{{\sin }^{2}}( \frac{\pi }{6}-\theta ) $ $ =\cos ( \frac{\pi }{6}+\theta +\frac{\pi }{6}-\theta )\cos ( \frac{\pi }{6}+\theta -\frac{\pi }{6}+\theta ) $ $ [\because {{\cos }^{2}}A-{{\sin }^{2}}B=\cos (A+B)\cos (A-B)] $ $ =\cos \frac{2\pi }{6}\cos 2\theta =\frac{1}{2}\cos 2\theta $ .
BETA
