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[29%" cos "20^(@)+cos^(2)40^(@)+cos^(2)8...

[29%" cos "20^(@)+cos^(2)40^(@)+cos^(2)80^(@)=],[[" 1) "(1)/(2)," 2) "1," 3) "(3)/(2)," 4) "2]]

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Cos^(2)20^(@)+Cos^(2)40^(@)+cos^(2)80^(@)=

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

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

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

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

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

cos 40^(@) + cos 80^(@) + cos 160^(@) =

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

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

Prove that: cos80^(@)+cos40^(@)-cos20^(@)=0