Problem:

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)

Back to the question paper.

*Related*

## One reply on “Sequence of tangents (I.S.I. B.Stat and B.Math 2017, subjective problem 1)”

[…] be written with an odd denominator. ……………………… Discussion […]

Google