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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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The value of $\cos 12^{\large\circ}+\cos 84^{\large\circ}+\cos 156^{\circ}+\cos 132^{\large\circ}$ is

$(a)\;\large\frac{1}{2}\qquad$$(b)\;1\qquad(c)\;-\large\frac{1}{2}$$\qquad(d)\;\large\frac{1}{8}$

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1 Answer

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$\cos 12^{\large\circ}+\cos 84^{\large\circ}+\cos 156^{\circ}+\cos 132^{\large\circ}=(\cos 132^{\large\circ}+\cos 12^{\large\circ})+(\cos 156^{\circ}+\cos 84^{\circ})$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;=(\cos 132^{\large \circ}+\cos 12^{\large \circ}+(\cos 156^{\large\circ}+\cos 84^{\large\circ})$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;=2\cos 72^{\large\circ}\cos 60^{\large\circ}+2\cos 120^{\circ}\cos 36^{\large\circ}$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;=\cos 72^{\large\circ}-\cos 36^{\large \circ}$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;=\large\frac{\sqrt 5-1}{4}-\frac{\sqrt 5+1}{4}$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;=\large\frac{-1}{2}$
Hence (c) is the correct answer.
answered Oct 8, 2013 by sreemathi.v
 

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