cos2x/(√2cos(x+π/4))=1/5,0
cos2x/根号2cos(x+π/4)=1/5,0
cos[2(π-x)]=-cos2x对不对?
cos^2 x=(cos2x-1)/2
已知sin x/2 -- 2cos x/2=0.(1)求tan x的值.(2)求 cos2x/(√2cos(π/4+x)
若cos(π/2+x)=4/5,则cos2x=
f(x)=(1+cos2x)/[4sin(pai/2+x)]-asin(x/2)cos(pai-x/
f(x)=cos²x·cos2x/1-cos²x(x∈0,π)的最小值是
求函数f(x)=cos平方x+cos2x的最大值?(1+cos2x)/2+cos2x 然后怎么计算?
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]
已知函数f(x)=(6cos^4x+5sin^2x-4)/cos2x
cos(x-π/2)=3/5.则cos2x=