The value of #sin20.sin40.sin60.sin80# is?

Question

63691 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-10T20:18:58

#sin20^circsin40^circsin60^circsin80^circ=3/16#

Let ,

#X=sin20^circsin40^circsin60^circsin80^circ#

#:.X=1/2(sin60^circsin20^circ)(2sin80^circsin40^circ)#

Using ,#"Product Identity"#

#2sinxsiny=cos(x-y)-cos(x+y),for ,x=80,y=40#

#:.X=1/2(sqrt3/2sin20^circ){cos(80-40)-cos(80+40)}#

#:.X=sqrt3/4sin20^circ{cos40^circ-cos120^circ}#

#:.X=sqrt3/4sin20^circ{cos40^circ+1/2}#

#:.X=sqrt3/8{2sin20^circcos40+sin20}#

Using ,#"Product Identity"#

#2sinxcosy=sin(x+y)+sin(x-y),for ,x=20,y=40#

#:.X=sqrt3/8{sin60+sin(-20)+sin20}#

#:.X=sqrt3/8{sqrt3/2-sin20^circ+sin20^circ}#

#:.X=sqrt3/8{sqrt3/2}#

#:.X=3/16#