作业帮sin(-1200°)-cos1290°+cos(-120)°+cos(-1020)°-sin(-1050°)+tan855°
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/29 03:06:20
sin(-1200°)-cos1290°+cos(-120)°+cos(-1020)°-sin(-1050°)+tan855°
sin(-1200°)-cos1290°+cos(-120)°+cos(-1020)°-sin(-1050°)+tan855°
sin(-1200°)-cos1290°+cos(-120)°+cos(-1020)°-sin(-1050°)+tan855°
sin(-1200°)=-sin(3*360° +120° )=-sin120°=-1/2
cos1290°=cos(4*360°-150°)=cos150°=-√3/2
cos(-120)°=cos120°=-1/2
cos(-1020)°=cos1020°=cos(3*360°-60°）=1/2
sin(-1050°)=sin(3*360°-30°）=-sin30°=-1/2
tan855°=tan135°=-1
sin(-1200°)-cos1290°+cos(-120)°+cos(-1020)°-sin(-1050°)+tan855°
=-1/2-（-√3/2）-1/2+1/2-（-1/2）+（-1）=（√3-2）/2
sin(-1200°)-cos1290°+cos(-120)°+cos(-1020)°-sin(-1050°)+tan855°
求sin(-1200°）×cos1290°+cos9-1020°)×sin（-1500°）+tan945°
求sin(-1200°）×cos1290°+cos9-1020°)×sin（-1500°）+tan95°
sin(-1200°)cos1290°+cos(-1020°）sin（-1050°)+tan855求值..
求sin(-1200°)×cos1290°+cos(-1020°)×sin(-1050°)的值求解求解吖~~~~
求sin(―1200°)·cos1290°+cos(―1020°)·sin(―1050°)+tan945°的值．
求sin﹙-1200°﹚·cos1290+cos﹙-1020°﹚·sin﹙-1050°﹚+tan945°的值
求音乐：sin( 1200度)cos1290度 cos( 1020度)in( 1050度)+tan495度
sin(10°)+sin(50°)-sin(70°) =sin(10°)+sin(50°)-sin(70°) =
sin(-1200°)的值是多少?
sin(18°)sin(54°)sin(18°)*sin(54°) 不用计算器
sin?°≈0.6773?
sin 15°是多少
sin²1°+sin²2°+sin²3°+.+sin²89
化简:sin^A*sin^A+sin(A-60°)*sin(A-60°)+sin(A+60°)*
sin(-1200°）.cos1209°+cos（-1020°)-sin(-1050°）+tan855°
计算：sin(-1200°)×cos1230°+cos(-1020°)sin(-1050°)+tan1305°
计算:sin方10°+sin方20°+sin方30°+sin方40°+sin方50°+sin方60°+sin方70°+sin方80°+sin方90°=