2sin^2x +cosx-1≥0怎么变成2cos^x-cos-1≤0
sin(x+1/2π)=cosx?还是cos-x
证明x∈(0,π/2),cos(cosx)>sin(sinx)
设(2cosx-sinx)(sin^2x+2cos^2x)=0,则(2cos^2x+sin^2x)/(1+tanx)等于
1、cos(x+B)*cos(x-B)=cosx-sin^2B
化简sin^4x/sinx-cosx - (sinx+cosx)cos^2x/tan^2x-1
证明成立:[cos(3x)-sin(3x)]/(cosx+sinx)=1-2sin(2x).
设x∈(0,π/2),则函数(sin²x+1/sin²)(cosx²+1/cos²
(1+2sinxcosx+cos^2x-sin^2x)+1这个是怎么到这步的[(cosx+sinx)(cosx+sinx
已知4sin^2x-cos^2x+3cosx=0求(cos2x-cos^2x)/(1-cot^2x)
2cos x (sin x -cos x)+1
lim(sin(x^2*cos(1/x)))/x怎么做?
(2sinx*cosx)/sin^2x+cos^2x=(2tanx)/tan^2x+1 怎么推导的?