计算:sin(-1200°)×cos1230°+cos(-1020°)sin(-1050°)+tan1305°
1个回答
展开全部
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
参考http://zhidao.baidu.com/question/120886616.html
=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
参考http://zhidao.baidu.com/question/120886616.html
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询