Question

已知数列{θn} {an}满足θ1=45°,且sinθn=an+1,tanθn=an,求数列an的通项公式

Answer 1

an = tanθn

a1=1

a(n+1) =sinθn

= ±an/√(1+(an)^2)

[a(n+1)]^2 = (an)^2/(1+(an)^2)

1/[a(n+1)]^2 = (1+(an)^2)/(an)^2

1/[a(n+1)]^2 - 1/(an)^2 =1

{ 1/(an)^2 } 是等差数列,d=1

1/(an)^2 - 1/(a1)^2 = n-1

1/(an)^2 = n

an = √(1/n)