^2 ( 6 + ) - ^2 ( 6 - ) = - Vidyadip

MCQ

${\cos ^2}\left( {\frac{\pi }{6} + \theta } \right) - {\sin ^2}\left( {\frac{\pi }{6} - \theta } \right) = $

✓
$\frac{1}{2}\cos 2\theta $
B
$0$
C
$ - \frac{1}{2}\cos 2\,\theta $
D
$\frac{1}{2}$

✓

Answer

Correct option: A.

$\frac{1}{2}\cos 2\theta $

a
(a) ${\cos ^2}\left( {\frac{\pi }{6} + \theta } \right) - {\sin ^2}\left( {\frac{\pi }{6} - \theta } \right)$

$ = \cos \left( {\frac{\pi }{6} + \theta + \frac{\pi }{6} - \theta } \right)\cos \left( {\frac{\pi }{6} + \theta - \frac{\pi }{6} + \theta } \right)$

$[ \because {\cos ^2}A - {\sin ^2}B = \cos (A + B)\cos (A - B)]$

$ = \cos \frac{{2\pi }}{6}\cos 2\theta = \frac{1}{2}\cos 2\theta $.

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