求值:arcsin(sin^2/2)rt
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/07 07:15:20
求值:arcsin(sin^2/2)rt
求值:arcsin(sin^2/2)
rt
求值:arcsin(sin^2/2)rt
因为√2/2属于[-π/2,π/2]
而arcsin的值域是[-π/2,π/2]
所以arcsin(sin√2/2)=√2/2
求值:arcsin(sin^2/2)rt
求值:sin(1/2arcsin(-12/13))
求值1.cos[arcsin(-4/5)-arctan(-3/4)]2.sin[1/2arcsin(-12/13)]
三角函数求值 sin(arcsin(1/2) + arccos (1/3))=?sin(arcsin(1/2) + arccos (1/3))=?
求值arctan( cot2),tan(1/2arccos4/5),cos[arcsin(-3/5)],arcsin(cos1/3)
求值sin[1/2arcsin(-3/5)]请标出反三角函数的范围.
求值arcsin(-√2/2)=
求值tan(arcsin√2/2)
求值:cos(arcsin√3/2)=
求值:arcsin(-√3/2)=
求值:arcsin(sin3π/4)rt
求值cos[arctan2+arcsin(-1/3)]rt
求值过程sin(arcsin(-1/3))
求值:arcsin√2/2+arccos1/2+arcsin√3/2+arccos√2/2=
求值 cos[arcsin(-根号3/2)] 3Q
求值:(1)arcsin(-√3/2)+arccos(-√2/2)(2)arcsin(√5/5)+aiccot3
arcsin(sinπ/3)=?arcsin(sinπ/3)的值等于 ( )A> π/3 B> 1/2请问我错在哪里?请大人把过程写上 是这样的arcsin[sin(π/3)]
arcsin[sin(19Π/12)]求值Π为3.1415926