Q) Let < a_n> be a sequence such that a₁ =1 and a_n+1 = cos a_n ; n>=1. If a = lim_n---∞ a_n then a belongs to the interval..?

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Given}: \\ {a}_1=1 \\ \mathrm{and} a_{n+1}=\cos a_n \\ \Rightarrow a_2=\cos a_1=\cos 1 \\ \Rightarrow a_3=\cos a_2=\cos \left(\cos 1\right) \\ . \\ . \\ . \\ . \\ . \\ . \\ {a}_n=\underset{n-1 times}{\underbrace{\cos \cos \cos ............\cos }\left(1\right)} \\ \mathrm{Given} \mathrm{that} \\ a=\underset{n\to \infty }{\lim }\underset{n-1 times}{\underbrace{\cos \cos \cos ............\cos }\left(1\right)} \\ \mathrm{Since} 0<1<\frac{\mathrm{\pi }}{2} \mathrm{and} \mathrm{we} \mathrm{know} \mathrm{that} \cos \mathrm{x} \mathrm{is} \mathrm{a} \mathrm{decreasing} \mathrm{function} \mathrm{in} 0<x<\frac{\mathrm{\pi }}{2}. \\ \mathrm{Now} \mathrm{for} \mathrm{a} \mathrm{decreasing} \mathrm{function}, \\ {x}_1>x_2 \\ \Rightarrow f\left(x_1\right)<f\left(x_2\right) \mathrm{i}.\mathrm{e}. \mathrm{if} \mathrm{you} \mathrm{apply} \mathrm{decreasing} \mathrm{function} \mathrm{on} \mathrm{both} \mathrm{sides}, \mathrm{sign} \mathrm{of} \mathrm{inequality} \mathrm{reverses}. \\ Now \\ 1>\cos 1 \\ \mathrm{Taking} \text{'}\cos \text{'} \mathrm{on} \mathrm{both} \mathrm{sides} \\ \cos 1<\cos \left(\cos 1\right) \\ \mathrm{Taking} \text{'}\cos \text{'} \mathrm{on} \mathrm{both} \mathrm{sides} \\ \cos \left(\cos 1\right)>\cos \cos \left(\cos 1\right) \\ \cos \cos \left(\cos 1\right)>\cos \cos \cos \left(\cos 1\right) \\ . \\ . \\ . \\ . \\ \mathrm{Here} \mathrm{if} \mathrm{you} \mathrm{observe} \mathrm{the} \mathrm{pattern} \mathrm{carefully}, \mathrm{the} \mathrm{value} \mathrm{wil} \mathrm{never} \mathrm{exceed} \cos \left(\cos 1\right) \mathrm{and} \mathrm{will} \mathrm{never} \mathrm{be} \mathrm{less} \mathrm{than} \cos 1 \mathrm{whatever} \mathrm{be} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{n}. \\ \mathrm{Hence} \\ \mathbf{cos}\mathbf{1}\mathbf{<}\mathbf{a}\mathbf{<}\mathbf{cos}\left(\mathbf{cos}\mathbf{ }\mathbf{1}\right)\mathbf{ }\mathbf{ }\left[\mathbf{Answer}\mathbf{ }\right] \end{gathered}$$

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