Suppose $x_1= an ^{-1} 2x_2x_3$ $\_\_\_\_$ are the real numbers satisfying $\sin \left(x_{n+1}-x

Question

Suppose $x_1=	an ^{-1} 2>x_2>x_3>$ $\_\_\_\_$ are the real numbers satisfying $\sin \left(x_{n+1}-x_night)+2^{-(n+1)} \cdot \sin x_n \cdot \sin x_{n+1}=0$ for all $n>1$ and the sequence is convergent and, $l=\lim _{n ightarrow \infty} x_n$, the value of $4 l$ is $\_\_\_\_$ .

Professor PC Thomas Classes · Answer

Here, $\sin(x_{n+1} - x_n) + 2^{-(n+1)} \sin x_n \cdot \sin x_{n+1} = 0$

$\Rightarrow \quad \cot x_{n+1} - \cot x_n = 2^{-(n+1)} \Rightarrow \quad \cot x_n - \cot x_{n-1} = 2^{-n}$

$\Rightarrow \quad \cot x_{n+1} - \cot x_1 = \frac{1}{2^2} + \frac{1}{2^3} + \dots\dots + \frac{1}{2^{n+1}} \Rightarrow \quad \cot x_{n+1} = \frac{1}{2} + \frac{1}{2^2} + \dots\dots \frac{1}{2^{n+1}}$

$\Rightarrow \quad \lim_{nightarrow\infty} \cot x_{n+1} = 1 \quad 	herefore \quad \lim_{nightarrow\infty} x_{n+1} = \frac{\pi}{4}$

$l = \frac{\pi}{4} \Rightarrow 4l = \pi$