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

Prove that cos 20^(@)cos40^(@)cos60^(@)cos80^(@)=(1)/(16)

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

Statement-1: The vlaue of cos20^(@)cos40^(@)cos60^(@)cos80^(@)is1/16. Statement-2: for any theta, cos thetacos(60^(@)-theta)cos(60^(@)+theta)=1/4cos3theta

Statement-1: The vlaue of cos20^(@)cos40^(@)cos60^(@)cos80^(@)is1/16. Statement-2: for any theta, cos thetacos(60^(@)-theta)cos(60^(@)+theta)=1/4cos3theta

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

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

Prove that cos 20^@ cos 40^@ cos60^@ cos80^@=1/16

cos 20^(@) cos 40^(@) cos 80^(@)=

Prove that cos 20^@cos 40^@cos 60^@cos 80^@=1/16 .

The value of cos 20^(@) cos 40^(@)cos 60^(@) cos 80^(@) is equal to