求证[1]sin θ+cos φ=2cos θ+φ/2 cos θ-φ/2 [2]tan(x/2+π/4)+tan(x/
求证:sin^2/(sin-cos) - (sin+cos)/(tan^2 -1) =sin+cos
证明(1-2sin x cos x )/(cos^2x-sin^2x)=(1-tan x)/(1+tan x)
已知sinθ+√3cosθ=2cos(θ-φ),求tanφ.(0
求证 1+2sinxcox/cos∧2x-sin∧2x=1+tanx/1-tanx tan∧2θ-sin∧2θ=tan∧
已知tan(θ+π/4)=-2,求cosθ平方+sinθcosθ-1
已知tan=2,求(cos x+sin x)/(cos x-sin x)+sin^2x
求证 (1—tan^2X)/(1+tan^2X)=cos^2X—sin^2X
函数f(x)=sin(2x+φ)+根号三cos(2x+θ)是奇函数,则tanθ等于
求证 (sinθ+cosθ-1)(sinθ-cosθ+1)) /sin2θ=tanθ/2
求证(sinθ+cosθ-1)(sinθ-cosθ+1)/sin2θ=tanθ/2
求证 tan(2π-X)sin(-2π-X)cos(6π-X)/ sin(X+3π/2)*cos(X+3π/2)=-ta
证明1-tan^2x/1+tan^2x=cos^2x-sin^2x