已知sin α=4/5,α属于(0,π/2)则cos(π-2α
1个回答
cos(π-2α )
=-cos(2a)
=2sin²a-1
=2*(16/25)-1
=7/25
相关问题
已知cosα=4/5,α属于(0,π/2),则sin(π+α)=
已知sin(2π-α)=[4/5],α∈(3π2,2π),则[sinα+cosα/sinα-cosα]等于( )
已知α属于(π/2,3π/2),tan(α-7π)=-3/4,则sinα+cosα=?
已知sinα+cosα=3根号5\5,α属于(0,π\4),sin(β-π\4)=3\5,β属于(π\4,π\2)
已知tanα(α+π/4)=2,α(π,(3π/2)),则(2sinα+cosα)/(3cosα-2sinα)=
已知sin(7π-α)-3cos(3π/2+α)=2,则(sin(π-α)+cos(π+α))/(sinα+cos(-α
已知sin(α+π/2)=-√5/5,α∈(0,π)求[cos2(π/4+α/2)-cos2(π/4-α/2)]/[si
已知sin(α+2/π)=-√5/5,α∈(0,π),求cos∧2(π/4+α/2)-cos∧2(π/4-α/2)/si
已知tan(π+α)=3,则2cos(π-α)-3sin(π+α)/4sin(α+π/2)+sin(2π-α)=
(2008•和平区三模)已知sin(2π-α)=[4/5],α∈(3π2,2π),则[sinα+cosα/sinα-co