सिद्ध कीजिए कि - `sin 20^(@)sin 40^(@)sin 60^(@)sin 80^(@)=(3)/(16)`.

Explanation

`L.H.S. = sin 20^(@)sin 40^(@)sin 60^(@)sin 80^(@)`
`rArr L.H.S. =(1)/(2)sin 60^(@).sin20^(@)[2 sin 40^(@)sin 80^(@)]`
`= (1)/(2)sin 60^(@)sin 20^(@)[cos (40^(@)-80^(@))-cos (80^(@)+40^(@))]`
`=(1)/(2)sin 60^(@)sin 20^(@)[cos 40^(@)-cos 120^(@)] " " [because cos (-theta)=cos theta]`
`=(1)/(2)sin 60^(@)sin 20^(@)[cos 40^(@)+cos 60^(@)]`
`=(1)/(2).(sqrt(3))/(2)sin 20^(@)[cos 40^(@)+(1)/(2)] " " [because cos 60^(@)=(1)/(2) " एवं " sin 60^(@)=(sqrt(3))/(2)]`
`=(sqrt(3))/(8)[2 sin 20^(@)cos 40^(@)+sin 20^(@)]`
`=(sqrt(3))/(8)[sin (40^(@)+20^(@))-sin (40^(@)-20^(@))+sin 20^(@)] [because 2 cos A sin B = sin (A+B)-sin (A-B)]`
`=(sqrt(3))/(8)[sin 60^(@)-sin 20^(@)+sin 20^(@)]`
`=(sqrt(3))/(8)xx sin 60^(@)`
`=(sqrt(3))/(8)xx(sqrt(3))/(2)=(3)/(16) = R.H.S. " "` इति सिद्धम्