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

1 Answer
Aug 10, 2018

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

Explanation:

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