计算:sin(-1200°)·cos1230°+cos(-1020°)sin(-1050°)+tan1305°
来源:学生作业帮 编辑:拍题作业网作业帮 分类:数学作业 时间:2024/05/06 10:55:58
计算:sin(-1200°)·cos1230°+cos(-1020°)sin(-1050°)+tan1305°
sin(-1200+360*4)*cos(1230-360*3)+cos(-1020+3*360)*-sin(-1050+360*3)+tan(1305-180*7)
=sin(240)*cos(150)+cos(60)*sin(30)+tan(45)
=[-sin(60)]*[-cos(30)]+cos(60)*sin(30)+tan(45)
=-((根号)3)/2*-((根号)3)/2+1/2*1/2+1
=3/4+1/4+1
=2
=sin(240)*cos(150)+cos(60)*sin(30)+tan(45)
=[-sin(60)]*[-cos(30)]+cos(60)*sin(30)+tan(45)
=-((根号)3)/2*-((根号)3)/2+1/2*1/2+1
=3/4+1/4+1
=2
计算:sin(-1200°)·cos1230°+cos(-1020°)sin(-1050°)+tan1305°
sin(-1200°).cos1209°+cos(-1020°)-sin(-1050°)+tan855°
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