三角函数合一变形(很简单的,ycosx-sinx=2y+1--合一变形：根号y^2+1cos(x+α)=2y+1tanα是-1/y还是1/y?

Question

三角函数合一变形(很简单的,
ycosx-sinx=2y+1--合一变形：根号y^2+1 cos(x+α)=2y+1
tanα是-1/y还是1/y?

Answer 1

ycosx-sinx=2y+1何以变形为[√(y²+1)] cos(x+α)=2y+1 ；tanα是-1/y还是1/y?
ycosx-sinx=[√(y²+1)]{[y/√(y²+1)]cosx-[1/√(y²+1)]sinx}=[√(y²+1)](cosxcosα-sinxsinα)
=[√(y²+1)]cos(x+α)=2y+1
其中cosα=y/√(y²+1),sinα=1/√(y²+1),tanα=sinα/cosα=1/y.