已知cos(5π/12-α)=m(lml≤1),求cos(π/12+α)和cos(11π/12-α)的值
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/27 05:01:01
已知cos(5π/12-α)=m(lml≤1),求cos(π/12+α)和cos(11π/12-α)的值已知cos(5π/12-α)=m(lml≤1),求cos(π/12+α)和cos(11π/12-
已知cos(5π/12-α)=m(lml≤1),求cos(π/12+α)和cos(11π/12-α)的值
已知cos(5π/12-α)=m(lml≤1),求cos(π/12+α)和cos(11π/12-α)的值
已知cos(5π/12-α)=m(lml≤1),求cos(π/12+α)和cos(11π/12-α)的值
π/12+α+(5π/12-α)=π/2,
11π/12-α-(5π/12-α)=π/2,
∴cos(π/12+α)=sin(5π/12-α)=土√(1-m^2),
cos(11π-α)=-sin(5π/12-α)=干√(1-m^2).
已知cos(5π/12-α)=m(lml≤1),求cos(π/12+α)和cos(11π/12-α)的值
已知lml=3,lml=5,m与n异号,求lm-nl的值
(1)已知sina=m(lml
若1/m- lml=1 则1/m+lml=?
若lml=2,lnl=5,且m>n,则m+n=?
若(m-1)x^lml=5是一元一次方程,求m的值
若(m-1)x^lml=5是一元一次方程,求m的值,
已知方程(m+ 2)x的lml-1次方=1是关于x的一元一次方程,则m=
已知(m-2)x的lml-1次方=1是一元一次方程,求m,用方程解,
若lml=l-3l,则m=多少
lml=lnl那么m和n的关系是( )
lml=5,lnl=3且mn>0求:(m+n)-(-lm-nl)的值
m,n互为相反数,m大于n,lml+lnl=1,求m.n的值
m,n互为相反数,m大于n,lml+lnl=1,求m.n的值
若lm—2l=2—m,lml=3,则m=()
如果(m+2)x^lml-1是一元一次方程,则m=( )
已知cos(π/6-α)=m(|m|≤1),求cos(5π/6+α),sin(2π/3-α)的值
已知sinα+cosα=m,(|m|