sin2θ=cos2θ+1,则cos2θ=?

sin2θ=cos2θ+1,则cos2θ=? 已知sin2/θ+cos2/θ=1/2,则cos2θ= 若sin2θ=cos2θ+1,则cos2θ= 1+sin2θ-cos2θ/1+sin2θ+cos2θ=tgθ 30分! 证明:1+sin2θ+cos2θ/1+ sinθ-cos2θ=tanθ 求证1+sin2θ -cos2θ/1+sinθ +cos2θ =tgθ 求证sinθ(1+cos2θ)=sin2θcos2θ sinθ+sin2θ/1+cosθ+cos2θ= (tanθ-1)/(2tanθ+1)=1,则cos2θ/(1+sin2θ)=? 若tanθ=2.则(sin2θ-cos2θ)/1+cos^2θ等于 若tan(π/4-θ)=3,则(cos2θ)/(1+sin2θ)等于? 速求!数学三角函数 已知sin2θ＝cos2θ＋1则cos2θ＝已知sin2θ＝cos2θ＋1则cos2θ＝速求答案! cotθ/(2cosθ+1)=1,求cos2θ和 cos2θ/(1+sin2θ) 化简sin2θ+sinθ/cos2θ+cosθ+1 若tan θ=3,求sin2θ -cos2θ tanθ=2 sin2θ+2cos2θ 已知sin2θ=a,cos2θ=b,0 f(x)=sin2θ+cos2θ最小正周期