若cos(π/2+x)=4/5,则cos2x=
若cos(π/2+x)=4/5,则cos2x=
cos[2(π-x)]=-cos2x对不对?
cos(x-π/2)=3/5.则cos2x=
cos2x/根号2cos(x+π/4)=1/5,0
cos^2 x=(cos2x-1)/2
已知函数f(x)=(6cos^4x+5sin^2x-4)/cos2x
x∈(0,π/4)cos(π/4+x)=5/13 则cos2x=
已知tanx=-4/3,试化简(2+cos2x)/[(根号2)·cos(x-π/4)-sin2x]
若f(cosx)=cos2x,且cosx-sinx=4/5,则f(sin2x/cos(x+派/4)=?
求证:sin(π/4+x)/sin(π/4-x)+cos(π/4+x)/(π/4-x)=2/cos2x
已知函数f(x)=(6cos^4x+5sin^2x-4)/cos2x 判断f(x)的奇偶性
f(x)=(1+cos2x)/[4sin(pai/2+x)]-asin(x/2)cos(pai-x/