m=(√5sinA+1)/(cosA+2)mcosA+2m=√5sinA+1√5sinA-mcosA=2m-1設(shè)f(A)=√5sinA-mcosA由輔角公式得：-√[(√5)^2+(-m)^2]≤f(A)≤√[(√5)^2+(-m)^2]即-√(5+m^2)≤2m-1≤√(5+m^2)(2m-1)^2≤5+m^24m^2-4m+1≤5+m^23m^...-√[(√5)^2+(-m)^2]≤f(A)≤√[(√5)^2+(-m)^2]可以幫我解釋下嗎？為什么F(A)要大于-√[(√5)^2+(-m)^2]小于)√[(√5)^2+(-m)^2]