为什么sin^2x=1/2(1-cos2x)

sin²α*sin²β+cos²α*cos²β-1/2cos2αcos2β=（1-cos2α）/2*（1-cos2β）/2+（1+cos2α）/2*（1+cos2β）/2-1/2cos2α*cos2β=1/4（1+cos2α*cos2β-cos2α-cos2β）+1/4（1+cos2α*cos2β+cos2α+cos2β）-1/2cos2α*cos2β=1/4+1/ 证明:1-cos2α/sinα=2sinα 1+cos2θ+2sin²θ=2 1-cos2θ/1+cos2θ=tan²θ 化简√(2+cos2+sin^2 1) cos2β= -(1/2),则sinβ= sinα-cosα=1/2 求cos2α 求证；1-2sin Xcos X/cos2 X-sin2 X=1-tan X/1+tan X 2sin^2σ*sin^2β+2cos^2σcos^2β=1+cos2σ*cos2β 求证：cos2αcos2β=1/2{cos2（α+β）+cos2(α-β)} sin^2(x-π/4)=[1-cos2（x-π/4）]/2=(1+sin2x)/2 函数f(x)=x^2-2sinθ*x+sinθ在x∈[0,1]上有最小值-1/4,求cos2θ的值 设f(x-1)=x2_2x+3,g(sin x)=cos2 x+sin x-1,设f(x),g(x)的表达式.设f(x-1)=x2-2x+3,g(sin x)=cos2 x+sin x-1,设f(x),g(x)的表达式 已知函数f(x)=x^2-2sinθ*x+sinθ在[0,1]上有最小值-1/4,求cos2θ的值. 若tan(π-α)=1/3,则cos2α/2sinαcosα+cos2α的值为 参数方程x=2+sin²θ y=-1+cos2θ化为普通方程是 ((1+cos2α)/(3sin2α))*((2sin^2(α))/cos2α)的化简 ((1+cos2α)/(3sin2α))*((2sin^2(α))/cos2α)的化简 急 ((1+cos2α)/(3sin2α))*((2sin^2(α))/cos2α)的化简