cos6^(0)*cos42^(@)*cos60^(@)*cos78^(@)=

Prove that cos6^(@)*cos42^(@)*cos66^(@)cos78^(@)=(1)/(16)

4cos6^(@)cos42^(@)cos60^(@)cos78^(@)=

cos6^(@) cos42^(@) cos66^(@) cos78^(@) =

A=cos 20^(@)cos 40^(@) cos60^(@)cos 80^(@) , B=cos6^(@)cos42^(@)cos66^(@)cos78^(@) and C=cos 36^(@)cos72^(@)cos 108^(@)cos 144^(@) then

cos20^(@)cos40^(@)cos60^(@)cos80^(@)=

cos42^(@)+cos78^(@)+cos162^(@)=

A=cos20^(@)cos40^(@)cos60^(@)cos80^(@),B=cos60cos42^(@)cos66^(@)cos78^(@) and C=cos36^(@)cos72^(@)cos108^(@)cos144^(@) then

cos20^(@).cos40^(@).cos60^(@).cos80^(@) is :

cos20^(@).cos40^(@).cos60^(@).cos80^(@)=1/16