Find-2-2-2-2-2-2-2-2-2-

Question Number 160445 by HongKing last updated on 29/Nov/21

Find :

\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2 + \sqrt{2}}}{2} \cdot \frac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2} \cdot \dots = ?

Commented by HongKing last updated on 29/Nov/21

Yes my dear Sir but how please

Commented by mr W last updated on 30/Nov/21

= \frac{2}{π}

Answered by mr W last updated on 30/Nov/21

a_{1} = \frac{\sqrt{2}}{2} = \sqrt{\frac{1}{2}}

a_{2} = \frac{\sqrt{2 + \sqrt{2}}}{2} = \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}}

\dots

a_{n} = \sqrt{\frac{1}{2} + \frac{1}{2} a_{n - 1}}

2 a_{n}^{2} - 1 = a_{n - 1}

s a y a_{n} = \cos θ_{n}

{2 \cos}^{2} θ_{n} - 1 = \cos θ_{n - 1}

\cos 2 θ_{n} = \cos θ_{n - 1}

2 θ_{n} = θ_{n - 1}

θ_{n} = \frac{θ_{n - 1}}{2} = \frac{θ_{n - 2}}{2^{2}} = \dots = \frac{θ_{1}}{2^{n - 1}}

a_{1} = \cos θ_{1} = \sqrt{\frac{1}{2}} \Rightarrow θ_{1} = \frac{π}{4} = \frac{π}{2^{2}}

θ_{n} = \frac{θ_{1}}{2^{n - 1}} = \frac{π}{2^{n + 1}}

a_{n} = \cos θ_{n} = \cos \frac{π}{2^{n + 1}}

a_{1} a_{2} \dots a_{n} = \cos \frac{π}{2^{2}} \cos \frac{π}{2^{3}} \dots \cos \frac{π}{2^{n + 1}}

2 \sin \frac{π}{2^{n + 1}} (a_{1} a_{2} \dots a_{n}) = \cos \frac{π}{2^{2}} \cos \frac{π}{2^{3}} \dots \sin \frac{π}{2^{n}}

2^{2} \sin \frac{π}{2^{n + 1}} (a_{1} a_{2} \dots a_{n}) = \cos \frac{π}{2^{2}} \cos \frac{π}{2^{3}} \dots \sin \frac{π}{2^{n - 1}}

\dots \dots

2^{n} \sin \frac{π}{2^{n + 1}} (a_{1} a_{2} \dots a_{n}) = \sin \frac{π}{2} = 1

a_{1} a_{2} \dots a_{n} = \frac{1}{2^{n} \sin \frac{π}{2^{n + 1}}}

a_{1} a_{2} \dots a_{n} = \frac{2}{π} \times \frac{\frac{π}{2^{n + 1}}}{\sin \frac{π}{2^{n + 1}}}

L H S = \lim_{n \to \infty} (a_{1} a_{2} \dots a_{n}) = \frac{2}{π}

Commented by HongKing last updated on 29/Nov/21

Perfecy my dear Sir thank you so much

Commented by Tawa11 last updated on 30/Nov/21

Great sir

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