1. The basic thing is to establish that the sequence has a limit.

 

Let cos_n1 represent the cosine function applied n times on the number 1

 

Now 1>cos(cos(1))

 

i.e. 1>cos₂1.

Hence,

 cos₃1>cos₁1 (as cos is a decreasing function in (0,pi)

cos₂1>cos₄1

cos₅1>cos₃1 and so on

 

So, you get that the sequence cos1, cos₃1, cos₅1,... is an increasing sequence, while the sequence cos₂1, cos₄1, cos₆1,... is a decreasing sequence. The sequences are bounded and hence the sequences converge.

 

Suppose they converge to different limits l₁ and l₂.

 

Then l₁ and l₂. are solutions to the equation x = cos(cos(x)) in the interval (0,1)

 

Let f(x) = x-cos(cos(x)).

f'(x) = 1-sin(sin(x)).sin(x)>0.i.e. it is a strictly increasing function. You can prove that it has a solution in (0,1). So that solution is unique.

 

Hence l₁ = l₂.

 

The limit is the solution to x = cosx, which comes to around 0.73.