Question

sin20° sin40° sin60° sin80°=3/16​

Answer 1

Answer:

sin20. sin 40. sin60. sin80  

=> sin60[sin20.sin40.sin80]  

=>√3/2[sin20.sin(60-20).sin(60+20)]  

=>√3/2[sin 3(20)/4]  

=>√3/2[sin 60/4]  

=>√3/2[√3/2*4]  

=>√3/2*√3/8  

=3/16  

 

hope it's help u

Answer 2

Answer:

since,sin60=√ 3/2

= √ 3/2( sin20sin40sin80)

=√ 3/2  ( sin20sin80sin40)

=√ 3/4 [(2sin20sin40)sin80]

on applying  [cos(A-B)-cos(A+B) =  2sinAsinB]

we get,

= √ 3/4    (cos20-cos60)   sin80    [since,cos(-a)=cosa]

= √ 3/ 4    (cos20sin80-cos60sin80)

= √ 3/8(2sin80cos20-sin80)

= √ 3/8   (sin100+sin60-sin80)

= √ 3/8   ( √ 3/2+sin100-sin80 )

= √ 3/8    ( √ 3/2+sin(180-80)-sin80 )

= √ 3/8( √ 3/2+sin80-sin80 )     [since,sin(180-a)=sina]

=  √ 3/8( √ 3/2)

=  3/16

HOPE IT HELPS YOU