Hmm, this looks familiar.

Calculus Level 5

Define a recurrence relation with starting value x0 x_0 such that 1<x0<1 -1 < x_0 < 1 and

xn+1=1+xn2. x_{n+1}=\sqrt{\frac{1+x_n}{2}}.

What is the value of

cos(1x02n=1xn)=? \cos\left(\frac{\sqrt{1-x_0^2}}{\prod\limits_{n=1}^{\infty}x_n}\right)=\,\,?

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