sin^2x 1=2cos^2x 求tanx和cos2x

求证:Sin^2 x / (sinx-cosx) - (sin x + cos x)/(ta求证:Sin^2 x / (sinx-cosx) - (sin x + cos x)/(tan^2-1) = sin x + cos x 已知tan=2,求(cos x+sin x)/(cos x-sin x)+sin^2x 已知sin(x)+3cos(x)=2,求sin(x)-cos(x)/sin(x)+cos(x)=? 2sin^2x - 5sin x cos x=3 cos^2x,求X, 已知:tan x=2 求:sin x+cos x/sin x-cos x=? Sin x-sin y=2/3 cos x-cos y=1/2 求cos(x-y) tan(x)=2求sin(x)cos(x)=? tan x=2,求 sin x * cos x=? f(sin x)=cos 2x +1 求 f(cos x) 求y=（1+cos x)/(sin x+cos x+2)的值域. 已知tanx=根号下2 求：1）（sin x+cos x）/（cos x -sin x）的值.2）2sin^2 x-sin x cos x+cos^2 x 的值 已知sin[(π/4)-x)]=5/13,0<x<π/4.求cos2x/(cosπ/4+已知sin[(π/4)-x)]=5/13,0<x<π/4.求cos2x/[cos(π/4)+x] 已知0<β<π/2<α<π,cos[α-(β/2)]=-1/9,sin[(α/2)-β]=2/3 求cos(α+β) 已知α,β∈(0,π)且tan(α-β)=1/2,ta y=1-sin(2x)/sin(x)-cos(x)求T.最值 sin(x+y)sin(x-y)=k,求cos^2x-cos^2y 已知-π/2＜x＜0,sin x+cos x=1/5求sin 2x+2 sin 怎么证明2(1-sin x)(1+cos x)=(1-sin x+cos x)^2 求证（cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2x 化简(sin x+cos x-1)(sin x-cos x+1)/sin 2x