求sin(-1200°)×cos1290°+cos(-1020°)×sin(-1050°)的值

问题描述:

求sin(-1200°)×cos1290°+cos(-1020°)×sin(-1050°)的值
求解求解吖~~~~

原式=sin(-1200+360*4)*cos(1230-360*3)+cos(-1020+3*360)*-sin(
-1050+360*3)
=sin(240)*cos(150)+cos(60)*sin(30)
=[-sin(60)]*[-cos(30)]+cos(60)*sin(30)
=-(√3)/2*-(√3)/2+1/2*1/2
=1