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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
0 votes

$\sin 47^{\circ}+\sin 61^{\circ}-\sin 11^{\circ}-\sin 25^{\circ}$ is equal to

$(a)\;\sin 36^{\large \circ}\qquad(b)\;\cos 36^{\large\circ}\qquad(c)\;\sin 7^{\large\circ}\qquad(d)\;\cos 7^{\large\circ}$

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1 Answer

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$\sin 47^{\circ}+\sin 61^{\circ}-\sin 11^{\circ}-\sin 25^{\circ}=2\sin54^{\large\circ}\cos 7^{\large\circ}-2\sin18^{\circ}\cos 7^{\large\circ }$
$\qquad\qquad\qquad\qquad\qquad\qquad\quad=\cos 7^{\large\circ}(\sin 54^{\large \circ}-\sin 18^{\large\circ})$
$\qquad\qquad\qquad\qquad\qquad\qquad\quad=\cos 7^{\large\circ}(\cos 36^{\large \circ}-\sin 18^{\large\circ})$
$\qquad\qquad\qquad\qquad\qquad\qquad\quad=\cos 7^{\large\circ}\bigg[\big(\large\frac{\sqrt 5+1}{4}\big)-\big(\large\frac{\sqrt 5-1}{4}\big)\bigg]$
$\qquad\qquad\qquad\qquad\qquad\qquad\quad=\cos 7^{\large\circ}$
Hence (d) is the correct answer.
answered Oct 8, 2013 by sreemathi.v
 
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