若sin2θ=cos2θ+1,则cos2θ=

sin2θ=cos2θ+1,则cos2θ=? 若sin2θ=cos2θ+1,则cos2θ= 已知sin2/θ+cos2/θ=1/2,则cos2θ= 1+sin2θ-cos2θ/1+sin2θ+cos2θ=tgθ 30分! 若tanθ=2.则(sin2θ-cos2θ)/1+cos^2θ等于 若tan(π/4-θ)=3,则(cos2θ)/(1+sin2θ)等于? 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= 若tan θ=3,求sin2θ -cos2θ (tanθ-1)/(2tanθ+1)=1,则cos2θ/(1+sin2θ)=? 若3sinθ=cosθ,则cos2θ+sin2θ的值等于 若tanθ=3,则sin2θ-cos2θ的值 若cotθ=3,则cos2θ-1/2sin2θ的值为若cotθ=3,则cos2θ-(1/2)sin2θ的值为过程 若复数z=sin2θ -i(1-cos2θ)是纯虚数,则θ为多少?应该是sin2θ=0,-(1-cos2θ)不等于0之后的交集怎么取? sin2α+sin2β-sin2αsin2β+cos2αcos2β=1 证明 θ为锐角,若sinθ+sin2θ=1,则cos2θ+cos^6θ的值为