求sin(-1200°)×cos1290°+cos(-1020°)×sin(-1050°)的值求解求解吖~~~~_其他解答

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

人气：376 ℃　时间：2020-06-19 19:05:20

解答

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