Trigonometric Identities Question 62
Question: $ {{\cos }^{2}}( \frac{\pi }{4}-\beta )-{{\sin }^{2}}( \alpha -\frac{\pi }{4} )= $
Options:
A) $ \sin (\alpha +\beta )\sin (\alpha -\beta ) $
B) $ \cos (\alpha +\beta )\cos (\alpha -\beta ) $
C) $ \sin (\alpha -\beta )\cos (\alpha +\beta ) $
D) $ \sin (\alpha +\beta )\cos (\alpha -\beta ) $
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Answer:
Correct Answer: D
Solution:
$ {{\cos }^{2}}( \frac{\pi }{4}-\beta )-{{\sin }^{2}}( \alpha -\frac{\pi }{4} ) $ $ =\cos ,( \frac{\pi }{4}-\beta +\alpha -\frac{\pi }{4} ),\cos ,( \frac{\pi }{4}-\beta -\alpha +\frac{\pi }{4} ), $ $ =\cos (\alpha -\beta )\cos ( \frac{\pi }{2}-\overline{\alpha +\beta } )=\cos (\alpha -\beta )\sin (\alpha +\beta ) $ .
BETA
