(sin2π/3+cos11π/6)tan7π/4=
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/31 00:55:50
(sin2π/3+cos11π/6)tan7π/4=
(sin2π/3+cos11π/6)tan7π/4=
(sin2π/3+cos11π/6)tan7π/4=
(sin2π/3+cos11π/6)tan7π/4
=[sin(π-π/3)+cos(2π-π/6)]*tan(2π-π/4)
=(sinπ/3+cosπ/6)*(-tanπ/4)
=[(根号3)/2+(根号3)/2]*(-1)
=-根号3
(sin2π/3+cos11π/6)tan7π/4=
求下列各三角函数值 1、cos(-60º) 2、sin2π/3 3、tan5π/4 4、cos11π/6
sin4/3π*cos11/6π*tan3/4π=
3(cos11π/6+isin11π/6) 解复数三角形
3(cos11π/6+isin11π/6) sin3π/5-icos3π/5 复数三角形式
sin(-π/6)+cos11π/3+tan(-5π/6)=急
cos11π/4+tan(-7π/6)+sin(-π)+cos(-5π/6)
(cos11π/6+isin11π/6)^9的代数形式是多少怎样计算?
sin2/3π=
sin-3π/4= cos11π/6= tan13π/4= sin7π/2= cos-3π/2= sin10π/3=
已知点(tan4π/3,cos11π/6) 是θ终边上一点,求tanθ,sinθ,cosθ
比较cos11π/8与sin10π/9的大小,
复数 -根号3+i的三角形式可以表示为A.2(cosπ/6+isinπ/6) B.2(cos11π/6+isinπ11π/6)C.2(cos5π/6+isin5π/6)D.2(cos11π/6-isinπ11π/6)选哪个???
求sin2(π/3-x)+sin2(π/6+x)的值
10.sin2(π/3 -x)+sin2(π/6 +x)=________ _; 望详解
化简sin2(a-π/6)+sin2(a+π/6)-sin2a
tan11π/*6cos(-25π/6)*sin(-23π/6)sin25π/6*cos11π/3+1/4tan(-29π/4)第一题应该是 tan11π/6*cos(-25π/6)*sin(-23π/6)
6(cos11π/6+isin11π/6)的代数形式是什么6[根3/(2)-(1/2i)]=3根号3-3i 为什么不是6[-根3/(2)-(1/2i)]=-3根号3-3i