2sin(π/4+a)=sinθ+cosθ,且2sin^2b=sin2θ,求sin2a+1/2cos2b
2sin(π/4+a)=sinθ+cosθ,且2sin^2b=sin2θ,求sin2a+1/2cos2b
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
为什么sin2θ+sinθ=2sinθcosθ+sinθ=sinθ(2cosθ+1)
求证sin平方a * sin平方b+cos平方a * cos平方b-1/2cos2a *cos2B=1/2
已知α∈(0,π/2),且2sinα-sinαcosα-3cosα=0.求[sin(α+π/4)]/[sin2α+cos
证明:2sinθ+sin2θ=4sinθ×cos^2(θ/2)
求证4sinθ(cosθ/2)^2=2sinθ+sin2θ
证明恒等式4sinθcos²θ/2=2sinθ+sin2θ
4sinθcos²θ/2=2sinθ+sin2θ
4sinΘcos²(θ/2)=2sinΘ+sin2Θ
若sin(π/4+α)=sinθ+cosθ,2sin^2β=sin2θ,求证:sin2θ+2cos2β=3