Sequence of tangents (I.S.I. B.Stat and B.Math 2017, subjective problem 1)
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Let the sequence \( { a_n} _{n \ge 1 } \) be defined by \(a_n = \tan n \theta \) where \( \tan \theta = 2 \). Show that for all n \( a_n \) is a rational number which can be written with an odd denominator.
Discussion:
This is simple induction.
The claim is true for \(n = 1\) clearly \( \tan 1\theta = \tan \theta = 2 = \frac {2}{1} =\) a rational number with an odd denominator.)
Assume the claim is true for n=k; that is suppose \( \tan k \theta = \frac{p}{2t+1} \) that is a rational number with odd denominator. (since the denominator is odd we can express it as 2t +1). Of course we know this to be valid for k = 1.
Next we show for n = k +1
\( \displaystyle { \tan (k+1) \theta = \tan (k \theta + \theta) \ = \frac { tan k \theta + \tan \theta }{ 1- \tan \theta \tan k \theta } \ = \frac { \frac {p}{2t +1} + 2 }{ 1 - 2 \frac{p}{2t+1} } \ = \frac { p+ 4t + 2} { 1 + 2t - 2p} \ = \frac {p+4t + 2} { 1 + 2(t-p) }} \)
Since p and t are integers, hence the ratio we found is rational. Denominator is \(1 + 2(t-p)\) (which is even + 1). Hence it is odd.
(proved)
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