Решите пожалуйста пример

Sin7гр15мин= $\sqrt{(1-cos15)/2}$ = $\sqrt{(1- \sqrt{(1-cos30)/2})/2 } = \sqrt{(1- \sqrt{(1- \sqrt{3 } /2)/2)}$ = $\sqrt{(1- \sqrt{(2- \sqrt{3)}/4 })/2 } = \sqrt{(1- \sqrt{(2- \sqrt{3} )} /2)/2} = \sqrt{(2- \sqrt{2- \sqrt{3} }) }$ /2=1/2 $\sqrt{2- \sqrt{2- \sqrt{3} } }$
sin75=sin(90-15)=cos15= $\sqrt{(1+cos30)/2} = \sqrt{(1+ \sqrt{3} /2)/2} = \sqrt{(2+ \sqrt{3} )/4} =1/2 \sqrt{2+ \sqrt{3} }$
2+log(2)sin7гр15мин-log(1/2)cos15гр=2+log(2)sin7гр15мин+log(2)cos15гр=
=
2+log(2)1/2+1/2log(2)(2- $\sqrt{2- \sqrt{3} } )+log(2)1/2+1/2log(2+ \sqrt{3} )=2-1-1+1/2log(2- \sqrt{2- \sqrt{3} })(2+ \sqrt{3})=1/2log(2)( 4-2 \sqrt{3} -2 \sqrt{2- \sqrt{3} } - \sqrt{6-3 \sqrt{3} } )$