求证:(sinα)^4+(cosα)^4=1-1\2*(sin2α)^2
求证:(sinα)^4+(cosα)^4=1-1\2*(sin2α)^2
求证:sin2α/(1+2sinα+cosα)=sinα+cosα-1
求证:2sinα+sin2α=[ 2 × (sinα)三次方 ]÷( 1-cosα )
求证:2sinα+sin2α=2sin^3α/(1-cosα)
三角恒等变换,急~求证(1-cosα+sinα )/(1+cosα+sinα)=tan(α/2) (3sin2α-4co
求证cos^8α-sin^8α 1/4sin2αsin4α=cos2α
求证2(sin2α+1)/1+sinα+cosα=tanα+1
若sin(π/4+α)=sinθ+cosθ,2sin^2β=sin2θ,求证:sin2θ+2cos2β=3
求证:sin2α/(1+sinα+cosα)=sinα+cosα-1
求证:sin2α/(1+sinα+cosα)=sinα+cosα-1
若2sin(π/4+θ)=sinθ+cosθ,2sin^2(β)=sin2θ,求证sin2α+1/2cos2β=0
已知3sin^2 α+2sin^2 β=1,3sin2α-2sin2β=0 求证cos(α+2β)=0