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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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The value of $\cos 12^{\large\circ}+\cos 84^{\large\circ}+\cos 156^{\large\circ}+\cos 132^{\large\circ}$ is

$\begin{array}{1 1}(A)\;\large\frac{1}{2}&(B)\;1\\(C)\;\large\frac{-1}{2}&(D)\;\large\frac{1}{8}\end{array} $

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1 Answer

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  • $\cos a+\cos b=2\cos\large\frac{(a+b)}{2}$$.\cos \large\frac{(a-b)}{2}$
$\cos 12^{\large\circ}+\cos 84^{\large\circ}+\cos 156^{\large\circ}+\cos 132^{\large\circ}$
$\cos 156^{\large\circ}+\cos 84^{\large\circ}+\cos 132^{\large\circ}+\cos 12^{\large\circ}$
$2\cos(\large\frac{156+84}{2})$$\cos(\large\frac{156-84}{2})$$+2\cos(\large\frac{132+12}{2})$$\cos(\large\frac{132-12}{2})$
$2\cos 120^{\large\circ}\cos 36^{\large\circ}+2\cos 72^{\large\circ}.\cos 60^{\large\circ}$
$2\times -\large\frac{1}{2}$$\cos 36^{\large\circ}+2\times \large\frac{1}{2}$$\cos 72^{\large\circ}$
$\cos 72^{\large\circ}-\cos 36^{\large\circ}$
$\large\frac{-1}{2}$
Hence (C) is the correct answer.
answered Apr 21, 2014 by sreemathi.v
 

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