求证:sinθ(1+cos2θ)=sin2θcosθsinθ*2(cosθ)^2点来架?可唔可以讲清楚D?唔该晒~

求证1+sin2θ -cos2θ/1+sinθ +cos2θ =tgθ 求证sinθ(1+cos2θ)=sin2θcos2θ 求证cos^4θ-sin^4θ=cos2θ 已知sinθ+cosθ=2sinα,sinθ·cosθ=sin²β,求证：2cos2α=cos2β. sinθ sinα cosθ成等差数列,sinθ sinβ cosθ为等比数列,求证2COS2α=cos2β 三角数列题：sinθ sinα cosθ成等差数列,sinθ sinβ cosθ为等比数列,求证2COS2α=cos2β ：求证:(1-2sinθcosθ)/(cos2θ-sin2θ)=(cos2θ-sin2θ)/(1+2sinθcosθ).求证:(1-2sinθcosθ)/(cos2θ-sin2θ)=(cos2θ-sin2θ)/(1+2sinθcosθ).cos2θ指的是cos^2θ~对的 没抄错 你认为哪里错了？ 若2sin(π/4+α)=sinθ+cosθ,2sin²β=sin2θ,求证:sin2α+1/2cos2β=0 若2sin(π/4+a）=sinθ+cosθ,2sin^2β=sin2θ,求证sin2a+(1/2)cos2β=0. 若2sin(π/4+a）=sinθ+cosθ,2sin^2β=sin2θ,求证sin2a+(1/2)cos2β=0 tan^2θ =2tan^2Ф+1 求证 cos2θ +sin^2Ф=0如题.sin^2a/(1-sin^2a)=2sin^2⊙/(1-sin^2⊙)+1sin^2a/(1-sin^2a+sin^2a)=(sin^2⊙+1)/(1-sin^2⊙+sin^2⊙+1)这是怎么来的。 若2sin（π/4+θ）=sinθ+cosθ,2sin^2（β）=sin2θ,求证sin2α+1/2cos2β=0sin2α = （1/2）cos2β 我知道了谢谢 若sinθ,sinα,cosθ成等差数列,sinθ sinβ cosθ为等比数列,求证2cos4α=cos2β sinθ+sin2θ/1+cosθ+cos2θ= 已知sin(π+θ)=-1/3则COS2θ 求证：sin2θ+sinθ/2cos2θ+2sin^2θ+cosθ=tanθ 证明:1+sin2θ+cos2θ/1+ sinθ-cos2θ=tanθ 1+cos2θ+2sin²θ=2 1-cos2θ/1+cos2θ=tan²θ