Question 1142055
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First use the identity  


    {{{sin(pi-alpha)}}} = {{{sin(alpha)}}},


which is valid for any angle {{{alpha}}}.  From this identity, you get


    {{{sin(4pi/5)}}} = {{{sin(pi/5)}}}.


It gives you


    {{{2*sin(4pi/5)*cos(pi/5)}}} = {{{2*sin(pi/5)*cos(pi/5)}}}.


Next use the identity  {{{2*sin(alpha)*cos(alpha)}}} = {{{sin(2*alpha)}}},  which is valid for any angle  {{{alpha}}}.  It gives you


    {{{2*sin(4pi/5)*cos(pi/5)}}} = {{{2*sin(pi/5)*cos(pi/5)}}} = {{{sin(2pi/5)}}}.


It is what has to be proved.
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