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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
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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
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答案和解析
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)
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)
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