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May 14, 2019

Trigonometric Substitution, ISI Entrance 2019 - Problem 6

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"] For all natural numbers\(n\), let          \(A_n=\sqrt{2-\sqrt{2+\sqrt{2+\cdots +\sqrt{2}}}}\)           (\( n\) many radicals) (a) Show that for \(n\ge 2,  A_n=2\sin \frac{π}{2^{n+1}}\). (b) Hence, or otherwise, evaluate the limit                              \(\displaystyle\lim _{n\to \infty} 2^n A_n\) .

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I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2019. Subjective Problem no. 6.
[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.22.4" open="off"]Trigonometric Substitution 

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.4" open="off"]7.5 out of 10

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Mathematical Olympiad Challenges by Titu Andreescu & Razvan Gelca. [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Start with hints

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Do you really need a hint? Try it first!

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For \(n\ge 2\), \(\sqrt{2+\sqrt{2+\cdots +\sqrt{2}}}\)       (\( n\) many radicals) =\(\sqrt{2+\sqrt{2+\cdots +\sqrt{2+0}}}\) =\(\sqrt{2+\sqrt{2+\cdots +\sqrt{2+2\cos \frac{π}{2}}}}\).   

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=\(\sqrt{2+\sqrt{2+\cdots +\sqrt{2(1+\cos \frac{π}{2})}}}\) =\(\sqrt{2+\sqrt{2+\cdots +\sqrt{4\cos^2 \frac{π}{2^2}}}}\) =\(\sqrt{2+\sqrt{2+\cdots +2\cos \frac{π}{2^2}}}\). (\(n-1\) many radicals)      ........    ...........     ..........      ...   .......  ...      .......     .........     ............    .....     ......... =\(2\cos \frac{π}{2^n}\)        \([n\ge 2]\).  

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Now \(A_n=\sqrt{2-2\cos \frac{π}{2^n}}\) \(\Rightarrow  A_n= \sqrt{2(1-\cos \frac{π}{2^n})}\) \(\Rightarrow  A_n= \sqrt{4\sin^2 \frac{π}{2^{n+1}}}\)    \(\Rightarrow  A_n= 2\sin \frac{π}{2^{n+1}}\). 

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Now \(\displaystyle\lim_{n\to \infty} 2^nA_n=\displaystyle\lim_{n\to \infty} 2^{n+1}\cdot \sin \frac{π}{2^{n+1}}\). Since , \(n \to \infty \) \(\Rightarrow 2^{n+1}\to \infty \) \(\Rightarrow \frac{π}{2^{n+1}}\to 0\) Let \(\frac{π}{2^{n+1}}=z \Rightarrow z \to 0\), when \(n \to \infty\). Therefore, \(\displaystyle\lim_{n\to \infty} 2^n A_n \)=\(\displaystyle\lim_{z \to 0}\frac{\sin z}{z}\cdot π =1\times π=π\). (Ans.)               

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Similar Problem

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