求证：cos²α+cos²（α+β）-2cosαcosβcos（α+β）=sin²2β

求证：cos²α+cos²（α+β）-2cosαcosβcos（α+β）=sin²2β
求证：cos²α+cos²（α+β）-2cosαcosβcos（α+β）=sin²2β

求证：cos²α+cos²（α+β）-2cosαcosβcos（α+β）=sin²2β
cos^2(α+β)
=(cosαcosβ-sinαsinβ)^2
=cos^2αcos^2β-2cosαcosβsinαsinβ+sin^2αsin^2β,
2cosαcosβcos(α+β),
=2cosαcosβ(cosαcosβ-sinαsinβ)
=2cos^2αcos^2β-2cosαcosβsinαsinβ,
——》
cos^2α+cos^2(α+β)-2cosαcosβcos(α+β)
=cos^2α+(cos^2αcos^2β-2cosαcosβsinαsinβ+sin^2αsin^2β)-(2cos^2αcos^2β-2cosαcosβsinαsinβ)
=cos^2α-cos^2αcos^2β+sin^2αsin^2β
=cos^2α(1-cos^2β)+sin^2αsin^2β
=cos^2αsin^2β+sin^2αsin^2β
=(cos^2α+sin^2α)sin^2β
=sin^2β,
命题得证.