求sin(-1200°)×cos1290°+cos9-1020°)×sin(-1500°)+tan945°
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/08/20 00:49:10
求sin(-1200°)×cos1290°+cos9-1020°)×sin(-1500°)+tan945°
sin(-1200°)×cos1290°+cos(-1020°)×sin(-1500°)+tan945°
=sin(1440-1200)×cos(1080+210)+cos(1080-60)×sin(-1440-60)+tan(720+225)
=sin240°×cos210°+cos60°×sin(-60°)+tan225°
=sin60°×cos30°-cos60sin60°+tan45°
=3/4-√3/4+1
=7/4-√3/4
=sin(1440-1200)×cos(1080+210)+cos(1080-60)×sin(-1440-60)+tan(720+225)
=sin240°×cos210°+cos60°×sin(-60°)+tan225°
=sin60°×cos30°-cos60sin60°+tan45°
=3/4-√3/4+1
=7/4-√3/4
请问 sin(-1200度)xcos1290度+cos(-1020度)xsin(-1050度)+tan945度 怎么算?
求sin(-1050°)三角函数值
化简COS8°-SIN17°SIN9°/COS9°
sin(-1200°).cos1209°+cos(-1020°)-sin(-1050°)+tan855°
化简(sin15°cos9°-cos66°)/(sin15°sin9°+sin66°)的结果是?
Sin(30°-a)=1/3,求Sina
计算:sin(-1200°)·cos1230°+cos(-1020°)sin(-1050°)+tan1305°
求值:(sin9°+sin6°cos15°)/(cos9°-sin6°sin15°)
计算:sin²1°+sin²2°+sin²3°...+sin²45°+sin
数列求和 sin²1°+sin²2°+sin²3°+.+sin²88°+sin&
高中三角函数题~sin(α+24°)=sin(α+66°) 求tanα?
已知y=3sin(x+20°)+5sin(x+80°),求该函数最大值