求sin﹙-1200°﹚·cos1290+cos﹙-1020°﹚·sin﹙-1050°﹚+tan945°的值

题目

求sin﹙-1200°﹚·cos1290+cos﹙-1020°﹚·sin﹙-1050°﹚+tan945°的值

答案

sin(-1200+360*4)*cos(1230-360*3)+cos(-1020+3*360)*-sin(-1050+360*3)+tan(45+180*5) =sin(240)*cos(150)+cos(60)*sin(30)+tan(45) =[-sin(60)]*[-cos(30)]+cos(60)*sin(30)+tan(45) =-(√3）/2*-(√3)/2+1/2*1/2+...