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## Homework Statement

The sequence {x

_{n}} is given by the recurrence relation

x

_{n}= cos(x

_{n-1})sin(x

_{n-2}) for n ≥ 2

and x

_{0}=2 and x

_{1}=1,4. Show by induction that 0 ≤ x

_{n}≤ 1 for all integers n ≥ 2.

## The Attempt at a Solution

We formulate a statement:

P

_{n}: 0 ≤ x

_{n}= cos(x

_{n-1})sin(x

_{n-2}) ≤ 1 for n ≥ 2

---------------------------------------------------------------

We assume that P

_{n}is true for n=k, where k ≥ 2:

A

_{k}: 0 ≤ x

_{k}= cos(x

_{k-1})sin(x

_{k-2}) ≤ 1 for k ≥ 2

We then set n=k+1:

x

_{k+1}= cos(x

_{k})sin(x

_{k-1})

We know from the assumption A

_{k}that:

0 ≤ x

_{k}≤ 1

This implies that:

0,54 ≤ cos(x

_{k}) ≤ 1 which again implies 0 < cos(x

_{k}) < 1

We then have to show:

0 ≤ sin(x

_{k-1}) ≤ 1

We know that sine to any angle is always equal to or less then 1. We therefore have to prove:

0 ≤ sin(x

_{k-1})

Since we assumed that 0 ≤ x

_{k}≤ 1, this must be true for x

_{k-1}as well. This means that if the assumption A

_{k}is true when n=k, its true when n=k+1 also. The last thing is to prove the base case:

x

_{2}= cos(1,4)sin(2) ≈ 0,155

0 ≤ 0,155 ≤ 1

P

_{n}is thus true.

Is this correct argumentation?