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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