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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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Connected Program at Cheenta

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Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

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