^2 76^o + ^2 16^o - 76^o16^o = - Vidyadip

MCQ

${\cos ^2}{76^o} + {\cos ^2}{16^o} - \cos {76^o}\cos {16^o} = $

A
$-1/4$
B
$1/2$
C
$0$
✓
$3/4$

✓

Answer

Correct option: D.

$3/4$

d
(d) ${\cos ^2}{76^o} + {\cos ^2}{16^o} - \cos {76^o}\cos {16^o}$

$ = \frac{1}{2}\left[ {1 + \cos {{152}^o} + 1 + \cos {{32}^o} - \cos {{92}^o} - \cos {{60}^o}} \right]$

$ = \frac{1}{2}\left[ {2 - \frac{1}{2} + \cos {{152}^o} + \cos {{32}^o} - \cos {{92}^o}} \right]$

$ = \frac{1}{2}\left[ {\frac{3}{2} + 2\cos {{92}^o}\cos {{60}^o} - \cos {{92}^o}} \right]$

$ = \frac{1}{2}\left[ {\frac{3}{2} + \cos {{92}^o} - \cos {{92}^o}} \right]$

$ = \frac{3}{4}$.

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