求证4sinθ(cosθ/2)^2=2sinθ+sin2θ
求证4sinθ(cosθ/2)^2=2sinθ+sin2θ
若sin(π/4+α)=sinθ+cosθ,2sin^2β=sin2θ,求证:sin2θ+2cos2β=3
若2sin(π4+α)=sin θ+cos θ,2sin2β=sin 2θ,求证:sin&
数学求证:sin2θ+sinθ/2cosθ+2sin²θ+cosθ=tanθ
求证 (sinθ+cosθ-1)(sinθ-cosθ+1)) /sin2θ=tanθ/2
求证(sinθ+cosθ-1)(sinθ-cosθ+1)/sin2θ=tanθ/2
:求证:(1-2sinθcosθ)/(cos2θ-sin2θ)=(cos2θ-sin2θ)/(1+2sinθcosθ).
为什么sin2θ+sinθ=2sinθcosθ+sinθ=sinθ(2cosθ+1)
若2sin(π/4+a)=sinθ+cosθ,2sin^2β=sin2θ,求证sin2a+(1/2)cos2β=0.
若2sin(π/4+a)=sinθ+cosθ,2sin^2β=sin2θ,求证sin2a+(1/2)cos2β=0
求证:sin2θ+sinθ/2cos2θ+2sin^2θ+cosθ=tanθ
证明:2sinθ+sin2θ=4sinθ×cos^2(θ/2)