Trigonometric Substitution, ISI Entrance 2019 - Problem 6
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 \(\displaystyle\lim _{n\to \infty} 2^n A_n\) .[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.3.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.3.4" hover_enabled="0"]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
[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.4" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!
[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]For \(n\ge 2\), \(\sqrt{2+\sqrt{2+\cdots +\sqrt{2}}}\) (\( n\) many radicals) =\(\sqrt{2+\sqrt{2+\cdots +\sqrt{2+0}}}\) =\(\sqrt{2+\sqrt{2+\cdots +\sqrt{2+2\cos \frac{π}{2}}}}\).
[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]=\(\sqrt{2+\sqrt{2+\cdots +\sqrt{2(1+\cos \frac{π}{2})}}}\) =\(\sqrt{2+\sqrt{2+\cdots +\sqrt{4\cos^2 \frac{π}{2^2}}}}\) =\(\sqrt{2+\sqrt{2+\cdots +2\cos \frac{π}{2^2}}}\). (\(n-1\) many radicals) ........ ........... .......... ... ....... ... ....... ......... ............ ..... ......... =\(2\cos \frac{π}{2^n}\) \([n\ge 2]\).
[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.4"]Now \(A_n=\sqrt{2-2\cos \frac{π}{2^n}}\) \(\Rightarrow A_n= \sqrt{2(1-\cos \frac{π}{2^n})}\) \(\Rightarrow A_n= \sqrt{4\sin^2 \frac{π}{2^{n+1}}}\) \(\Rightarrow A_n= 2\sin \frac{π}{2^{n+1}}\).
[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]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.)
[/et_pb_tab][/et_pb_tabs][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" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]Connected Program at Cheenta
[/et_pb_text][et_pb_blurb title="I.S.I. & C.M.I. Entrance Program" image="https://cheenta.com/wp-content/uploads/2018/03/ISI.png" _builder_version="3.22.4" header_level="h1" header_font="||||||||" header_text_color="#e02b20" header_font_size="50px" body_font="||||||||"]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.
[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/isicmientrance/" button_text="Learn More" button_alignment="center" _builder_version="3.22.4" custom_button="on" button_text_color="#ffffff" button_bg_color="#e02b20" button_border_color="#e02b20" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][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="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]