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[" 62."cos20^(@)cos40^(@)cos60^(@)cos80^...

[" 62."cos20^(@)cos40^(@)cos60^(@)cos80^(@)=],[[" 1) "1/2," 2) "1/4," 3) "1/8," 4) "1/16]]

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

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

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

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

cos20^@cos40^@cos80^@=

cos40^(@)+cos 80^(@)+cos160^(@)+cos240^(@) =

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

cos 20^(@)* cos 40^(@) * cos 60^(@) * cos 80^(@) =

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