Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 179917 by mnjuly1970 last updated on 04/Nov/22

Answered by a.lgnaoui last updated on 06/Nov/22

$$\frac{1}{4^n\cos {}^2\left(\frac{\pi }{2^{n+2}}\right)}=\frac{1}{4^n}(1+\tan {}^2\left(\frac{\pi }{2^{n+2}}\right)$$$$\tan \left(x\right)\cong x+\frac{x^3}{3}+\frac{2}{15}{x}^5{x}^2{(1+\frac{x^2}{3})}^2<{\tan }^2\left(x\right)<x^2{(1+\frac{x^2}{3}+\frac{2}{15}{x}^4)}^2$$$$x^2+\frac{2}{3}{x}^4+\frac{x^6}{9}<{\tan }^2\left(x\right)<x^2+\frac{2x^4}{3}+\frac{x^6}{9}+\frac{2x^6}{15}$$$$\frac{1}{4^n}[(1+{\left(\frac{\pi }{2^{n+2}}\right)}^2+\frac{2}{3}{\left(\frac{\pi }{2^{n+2}}\right)}^4+{\left(\frac{\pi }{2^{n+2}}\right)}^6]<\frac{1}{4^n}[1+\tan {}^2\left(\frac{\pi }{2^{n+2}}\right)]<\frac{1}{4^n}[1+{\left(\frac{\pi }{2^{n+2}}\right)}^2+\frac{2}{3}{\left(\frac{\pi }{2^{n+2}}\right)}^4+\frac{11}{45}{\left({}^6\frac{\pi }{2^{n+2}}\right)}^6]$$$$$$$$\frac{1}{2^{2n}}+\frac{{\pi }^2}{2^{4n+4}}+\frac{2^4{\pi }^4}{3\times 2^{6n+8}}<S<\frac{1}{2^{2n}}+\frac{{\pi }^2}{2^{4n+4}}+\frac{2}{3}\times \frac{{\pi }^4}{2^{6n+8}}+\left(\frac{11}{45}\times \frac{{\pi }^6}{2^{8n+12}}\right)$$$$\Rightarrow {\lim }_{\mathrm{n}\to \mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{4^n}\times \frac{1}{\cos {}^2\left(\frac{\pi }{2^{n+2}}\right)}={\lim }_{\mathrm{n}\to \mathrm{\infty }}\frac{1}{2^{2n}}=0$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com