Question

Sin20. Sin40. Sin60. Sin80=3/16

Answer 1

hey friend!!
here is your answer,


sin(20) sin(40) sin (60) sin (80) 

substitute sin(60) = √3 /2 

√3/2 [ sin(20) sin(40) sin(80) ] 

= (√3/2) sin(20) [ sin(40) sin(80) ] 

use the formula sin A sin B = (1/2) [ cos(A - B) - cos(A + B) ] 

= √3/2 sin(20) (1/2)[ cos(40) - cos(120) ] 

= √3/4 sin(20) [ cos(40) + cos(60) ] 

= √3/4 sin(20) [ cos(40) + 1/2 ] 

= √3/4 sin(20)cos(40) + (√3/8) sin(20) 

use the formula sin A cos B = 1/2 [ sin(A + B) + sin(A - B) ] 

= (√3/4)(1/2) [ sin(60) + sin(-20) ]+ (√3/8)sin(20) 

= (√3/8) [ (√3 / 2) - sin(20) ]+ (√3/8)sin(20) 

Answer 2

LHS = sin20 . sin40 . √3/2 .sin80 

= √3/4 . sin20(2sin40.sin80)

= √3/4 . sin20(cos40 - cos120) 

= √3/4 . sin20(cos40+1/2)

= √3/8 . sin20(2cos40+1)

= √3/8 .(2cos40.sin20 + sin20)

= √3/8 . (sin60 - sin20 + sin20) 

= √3/8 .sin60

= √3/8 . √3/2

= 3/16 = RHS