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Property 6

lf xn+1=√1+...

Question

lf $x_{n + 1} = \sqrt{\frac{1 + x_{n}}{2}}$ , then $cos ⎡ ⎢ ⎢ ⎣ \frac{\sqrt{1 - x_{0}^{2}}}{x_{1} x_{2} x_{3} \dots . \infty} ⎤ ⎥ ⎥ ⎦ (- 1 < x_{0} < 1)$ is equal to

A

- 1

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B

1

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C

x_{0}

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D

\frac{1}{x_{0}}

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Solution

The correct option is C $x_{0}$
Let $L = cos ⎡ ⎢ ⎢ ⎣ \frac{\sqrt{1 - x_{0}^{2}}}{x_{1} x_{2} x_{3} \dots . \infty} ⎤ ⎥ ⎥ ⎦ = lim n \to \infty cos ⎡ ⎢ ⎢ ⎣ \frac{\sqrt{1 - x_{0}^{2}}}{x_{1} x_{2} x_{3} \dots x_{n}} ⎤ ⎥ ⎥ ⎦$
Put $x_{0} = cos θ \Rightarrow x_{1} = \sqrt{\frac{1 + cos θ}{2}} = cos \frac{θ}{2}$
$\Rightarrow x_{2} = \sqrt{\frac{1 + cos \frac{θ}{2}}{2}} = cos \frac{θ}{2^{2}}$
Similarly $x_{n} = cos \frac{θ}{2^{n}}$
$∴ L = lim n \to \infty cos ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{\sqrt{1 - {cos}^{2} θ}}{cos \frac{θ}{2} cos \frac{θ}{2^{2}} . . . . . . . . . . . cos \frac{θ}{2^{n}}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
$= lim n \to \infty cos ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{sin θ}{cos \frac{θ}{2^{n}} cos \frac{θ}{2^{n - 1}} . . . . . . . . . . . cos \frac{θ}{2}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
$= lim n \to \infty cos ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{2 sin \frac{θ}{2^{n}} sin θ}{2 sin \frac{θ}{2^{n}} cos \frac{θ}{2^{n}} cos \frac{θ}{2^{n - 1}} . . . . . . . . . . . cos \frac{θ}{2}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
$= lim n \to \infty cos ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{2 sin \frac{θ}{2^{n}} sin θ}{sin \frac{θ}{2^{n - 1}} cos \frac{θ}{2^{n - 1}} . . . . . . . . . . . cos \frac{θ}{2}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦, [∵ 2 sin x cos x = sin 2 x]$
.........................................
.........................................
doing it $n$ times
$= lim n \to \infty cos ⎡ ⎢ ⎢ ⎢ ⎣ \frac{2^{n} sin \frac{θ}{2^{n}} sin θ}{sin θ} ⎤ ⎥ ⎥ ⎥ ⎦ = lim n \to \infty cos [2^{n} sin \frac{θ}{2^{n}}]$
$= lim n \to \infty cos ⎡ ⎢ ⎢ ⎢ ⎣ θ \frac{sin \frac{θ}{2^{n}}}{\frac{θ}{2^{n}}} ⎤ ⎥ ⎥ ⎥ ⎦ = cos θ = x_{0}$

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