设sin(π/4+θ)=1/3,则sin2θ=
设sin(π/4+θ)=1/3,则sin2θ=
sin(α+π/4)=1/3,则sin2α
若sin(π/4+α)=sinθ+cosθ,2sin^2β=sin2θ,求证:sin2θ+2cos2β=3
已知sin(θ+π/4)=1/3,求sin2θ
已知sinθ=4/5,且sinθ+cosθ>1,sin2θ=
已知 sinθ =4/5,且 sinθ-cosθ>1,则 sin2θ=
sinθ+sin2θ/1+cosθ+cos2θ=
急.设f(θ)=[2cos2θ+sin2(2π-θ)+sin(π/2+θ)-3]/[2+2cos2(π+θ)+cos(-
若2sin(π/4+θ)=sinθ+cosθ,2sin^2(β)=sin2θ,求证sin2α+1/2cos2β=0
设sin(π+α)=4/5,α∈(π/2,3π/2),则sin2α-【cos(a/2)】^2的值为----
证明 (2sinα-sin2α)/(2sinα+sin2α)=tan²(θ/2)
若2sin(π/4+a)=sinθ+cosθ,2sin^2β=sin2θ,求证sin2a+(1/2)cos2β=0.