x∈(0,π/4)cos(π/4+x)=5/13 则cos2x=
x∈(0,π/4)cos(π/4+x)=5/13 则cos2x=
f(x)=cos²x·cos2x/1-cos²x(x∈0,π)的最小值是
已知cos(π/4+x)=5/13,0<x<π/4,求sin(π/4-x)/cos2x
已知sin(π/4-x)=5/13,x属于(0,π/4),求(cos2x)/cos[(π/4)+x]
已知sin[(π/4)-x)]=5/13,0<x<π/4.求cos2x/(cosπ/4+
函数y=cos2x*cos(2x+π/6)(0≤x≤π/4)的最大值为
若cos(π/2+x)=4/5,则cos2x=
cos2x/根号2cos(x+π/4)=1/5,0
cos[2(π-x)]=-cos2x对不对?
1.y=cos^4x+sin^4x 求周期 2.y=(sin2x+sin(2x+π/3))/( cos2x+cos(2x
已知sin x/2 -- 2cos x/2=0.(1)求tan x的值.(2)求 cos2x/(√2cos(π/4+x)
已知tan(x-π/4)=2 求(sin2x+cos2x)/(2cos²x-3sin2x-1)