设cos(α-β/2)=-3/5,sin(α/2-β)=2/3,且π/2
设α β γ∈(0,π/2) ,且 sinβ+sinγ=sinα,cosα+cosγ=cosβ,则β—α 等于
sinα=sqr(2)sinβ,sqr(3)cosα=sqr(2)cosβ,且α,β均为三角形内角,求sinα,sinβ
设α为第一象限的角,且sinα=3/5,求(sin(π+α)+cos(5π-α))/(sin(α-6π)+2cos(2π
设cos(a-β/2)=-1/9,sin(a/2-β)=2/3,且π/2
设f(α)=2sinαcosα+cosα/1+sin²α+cos(3π/2+α)-sin²(π/2+
已知α,β属于(0,π/2)且sinβ=cos (α+β)sinα,
已知α β为锐角 且sinα-sinβ=-1/2 cosα-cosβ=1/3 则cos(α-β)
已知α∈(0,π/2),且2sinα-sinαcosα-3cosα=0.求[sin(α+π/4)]/[sin2α+cos
sinα+cosα=1/5,且π/2
已知sin(3π-α)=根号2*sin(π-β),根号3*cos(-α)=-根号2*cos(π+β),且α、β∈(0,π
α属于(0,π/2)且2sin²α-sinαcosα-3cos²α=0求sin(α+π/4)/sin
cos(π/2-α)=根号2cos(3/2π+β),根号3sin(3π/2-α)=-根号2sin(π/2+β)且0