PI in ].0 ,4.[ by SIN_COS:def 32;
then A1: 0 < PI by XXREAL_1:4;
then PI / 4 < PI / 1 by XREAL_1:78;
then A2: PI / 4 < PI + ((2 * PI ) * 0 ) ;
0 / 4 < PI / 4 by A1, XREAL_1:76;
hence sin (PI / 4) > 0 by A2, SIN_COS6:11; :: thesis: verum