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Ask iit jee aieee pet cbse icse state board community Discussion Response Post to: A sine problem
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bhargavi (67)

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 sin(/14)sin(3/14)sin(5/14)= cos(3/7)cos(2/7)cos(/7)
put   /7  as  t,then  cos3t=-cos4t
hence     cos3tcos2tcost=(-1)cos4tcos2tcost=[(-1)/2sint]*2sintcostcos2tcos4t
                                                                 =[(-1)/2sint]*sin2tcos2tcos4t
                                                                 =[(-1)/4sint]*sin4tcos4t
                                                                =[(-1)/8sint]*sin8t
                                                               =(-1)(sin8t/8sint)=(-1)(-1/8)=1/8
since   sin8/7=sin(+t)=(-1)sintsin8t=-sint
hence   answer  is   1/8
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