tan[arcsin(-3/5)]=怎么考虑
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/25 09:46:10
tan[arcsin(-3/5)]=怎么考虑tan[arcsin(-3/5)]=怎么考虑tan[arcsin(-3/5)]=怎么考虑设x=arcsin(-3/5),sin(x)=-3/5,(x为负角)
tan[arcsin(-3/5)]=怎么考虑
tan[arcsin(-3/5)]=
怎么考虑
tan[arcsin(-3/5)]=怎么考虑
设 x = arcsin(-3/5),
sin(x) = -3/5,(x为负角)
则tan[arcsin(-3/5)]=tan(x) = -3/4
-3/4
令arcsin(-3/5)=A,即sinA=-3/5(A为第三或第四象限角)
故tan[arcsin(-3/5)]=tanA=±3/4。
计算器-3/4
-3/4
tan[arcsin(-3/5)]=怎么考虑
tan[arcsin(-1/3)]
求值arctan( cot2),tan(1/2arccos4/5),cos[arcsin(-3/5)],arcsin(cos1/3)
计算 tan【arcsin(-1/3)】=?cos(arcsin3/5)=?
tan(arcsin(1/3)+arccos(-1/5))求值
arcsin(π/3)=
arcsin(cos4)=?怎么算?
y=(arcsin(x/3))^5的导数怎么求啊?
y=arcsin(x-3)的定义域怎么求?
arcsin(-√3)=
arcsin(sin4π/3)=
arcsin(sin4*3.14/3)=?
sec(arcsin(3/4))怎么解答
sin cos tan cot arcsin arccos这些是怎么转换呀?
tan{arcsin[cosπ/4]+arccos[sinπ/3]+arctan√3}
求值tan(arcsin√2/2)
化简sec(2arcsin(tan(arccotx))
1.化简:sin²1°+sin²2°+sin²3°+……+sin²89°=?2.如果f(sinx)=cos2x,那么f(cosx)等于?3.对于反三角函数式arccos5π/4,arcsin(log3 4),arcsin(根号2-1)²,arcsin(tanπ/3),有意义的式子的个数为?4.已