已知sinα与sinβ分别是sinθ与cosθ的等差中项与等比数列的中项,求证:2cos2α=cos2β=2{cos(π/4 +θ)}^2

因为2sinα=sinθ+cosθ,(sinβ)^2=sinθcosθ,所以(2sinα)^2=1+2sinθcosθ=1+2(sinβ)^2,即4(sinα)^2-2=2(sinβ)^2-1,所以2cos2α=cos2β.2{cos(π/4 +θ)}^2=2{cosπ/4cosθ-sinπ/4sinθ}^2=2{根号2/2(cosθ-s...