cos2α=cos^2(α)-sin^2(α)?怎么出来的?
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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/
cosα=2sinα则cos2α=
sinα-cosα=1/2 求cos2α
求证sin(α+β)+2cos(α-β)cosαcosβ=cos2α+cos2β
(2sinα+cosα)/(sinα-3cosα)=-5 求cos2α
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
sinαcosαcos2α=
2sinα=cosα,求cos2α+sin2α+1/cos^2 α
化简:2sinα^2cosα-sin(3/2π-α)cos2α=
cos2α/[sin(α-π/4)]=-根号2 /2,cosα+sinα=?
化简(sinα-cosα)^2-1/-cos2α
化简(sinα)^2*(sinβ)^2+(cosα)^2(cosβ)^2-1/2cos2αcos2β
化简:sin^2αsin^2β+cos^2αcos^2β-1/2cos2αcos2β
化解sin^2αsin^2β+cos^2αcos^2β-1/2cos2αcos2β
化简:sin²αsin²β+cos²αcos²β-1/2cos2αcos2β
化简 sin²αsin²β+cos²αcos²β-1/2(cos2αcos2β)