已知sinα=asinβ bcosα=acosβ
问题描述：
已知sinα=asinβ bcosα=acosβ
α、β为锐角
求证 cosα=根号下[(a²-1)/(b²-1)]
答
证明：sinα=asinβ,bcosα=acosβ,(sinα)^2=a^2(sinβ)^2,b^2(cosα)^2=a^2(cosβ)^2 两式相加,1-(cosα)^2+b^2(coaα)^2=a^2 (cosα)^2=(a^2-1)/(b^2-1) (b^2-1≠0) α是锐角,cosα=√(a^2-1)/(b^2-1)...