Đặt \(36^0= x\) ta có:
\(\eqalign{
& sin3x{\rm{ }} = {\rm{ }}sin{\rm{ }}\left( {{{180}^0} - 3x} \right) = sin2x \cr
& \Leftrightarrow 3\sin x - 4{\sin ^3}x = 2\sin x\cos x \cr
& \Leftrightarrow 3 - 4(1 - {\cos ^2}x) = 2{\mathop{\rm cosx}\nolimits} \cr
& \Leftrightarrow 4co{s^2}x - 2\cos x - 1 = 0 \cr
& \Rightarrow {\mathop{\rm cosx}\nolimits} = \cos {36^0} = {{1 + \sqrt 5 } \over 4} \cr} \)
Vậy : \(4(cos{24^0} + \cos {48^0} - \cos {84^0} - \cos {12^0}) \)\(= 2(1 + \sqrt 5 )\sqrt {{{3 - \sqrt 5 } \over 8}} = 2\)
Thay \(\cos {36^0} = {{1 + \sqrt 5 } \over 4}\) ta được: \(\tan {9^0} - \tan {63^0} + \tan {81^0} - \tan {27^0} \)\(= 4\)