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 48066 by rahul 19 last updated on 18/Nov/18

$$\sqrt{\frac{1}{2}}.\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}.\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}}......\mathrm{\infty }=?$$

Commented by MJS last updated on 18/Nov/18

$$=\frac{2}{\pi }$$$$\mathrm{approximated}\mathrm{it}\mathrm{but}\mathrm{I}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{prove}\mathrm{it}$$

Commented by maxmathsup by imad last updated on 19/Nov/18

$$wehave\sqrt{\frac{1}{2}}=cos\left(\frac{\pi }{4}\right),\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}=\sqrt{\frac{1+cos\left(\frac{\pi }{4}\right)}{2}}=cos\left(\frac{\pi }{8}\right)...$$$$S_n=cos\left(\frac{\pi }{4}\right)cos\left(\frac{\pi }{8}\right).cos\left(\frac{\pi }{16}\right)...cos\left(\frac{\pi }{2^n}\right)=\prod _{k=2}^{n}cos\left(\frac{\pi }{2^k}\right)$$$$W_n=\prod _{k=2}^{n}sin\left(\frac{\pi }{2^k}\right)\Rightarrow S_n.W_n=\prod _{k=2}^{n}cos\left(\frac{\pi }{2^k}\right)sin\left(\frac{\pi }{2^k}\right)$$$$=\frac{1}{2^{n-1}}\prod _{k=2}^{n}sin\left(\frac{\pi }{2^{k-1}}\right)=\frac{1}{2^{n-1}}\prod _{k=2}^n\prod _{k=2}^{n}sin\left(\frac{\pi }{2^{k-1}}\right)$$$$=\frac{1}{2^{n-1}}\prod _{k=2}^n\prod _{k=1}^{n-1}sin\left(\frac{\pi }{2^k}\right)=\frac{1}{2^{n-1}}\prod _{k=2}^n\frac{\prod _{k=2}^{n}sin\left(\frac{\pi }{2^k}\right)}{sin\left(\frac{\pi }{2n}\right)}=\frac{1}{sin\left(\frac{\pi }{2^n}\right)}\frac{1}{2^{n-1}}.W_n\Rightarrow$$$$$$$$S_n=\frac{1}{2^{n-1}sin\left(\frac{\pi }{2^n}\right)}=\frac{2}{2^{n}sin\left(\frac{\pi }{2^n}\right)}\sim \frac{2}{2^n.\frac{\pi }{2^n}}(n\to +\mathrm{\infty })\Rightarrow {lim}_{n\to +\mathrm{\infty }}{S}_n=\frac{2}{\pi }.$$

Commented by rahul 19 last updated on 20/Nov/18

thank you prof Abdo����

Commented by Abdo msup. last updated on 20/Nov/18

$$youarewelcomesir.$$

Answered by arcana last updated on 19/Nov/18

$$cos\left(\theta \right)=\sqrt{\frac{1}{2}}$$$$\sqrt{\frac{1}{2}}\cdot \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}=cos\left(\theta \right)\sqrt{\frac{1+cos\left(\theta \right)}{2}}=cos\left(\theta \right)cos\left(\theta /2\right)$$$$cos\left(\theta \right)cos\left(\theta /2\right)\sqrt{\frac{1+cos\left(\theta /2\right)}{2}}=cos\left(\theta \right)cos\left(\theta /2\right)cos\left(\theta /4\right)$$$$\mathrm{asi}\mathrm{podemos}\mathrm{escribirlo}\mathrm{como}$$$$cos\left(\theta \right)\cdot cos\left(\theta /2\right)\cdot cos\left(\theta /4\right)\cdot ...=\underset{i=0}{\prod ^{\mathrm{\infty }}}cos\left(\frac{\theta }{2^i}\right)$$$$\mathrm{si}\mathrm{multiplicamos}\mathrm{y}\mathrm{dividimos}\mathrm{por}sin\left(\frac{\theta }{2^i}\right)$$$$\frac{cos\left(\theta \right)\cdot cos\left(\theta /2\right)\cdot cos\left(\theta /4\right)\cdot ...\cdot \frac{1}{2}2cos\left(\theta /2^i\right)\cdot sin\left(\theta /2^i\right)}{sin\left(\theta /2^i\right)}$$$$\mathrm{usando}sin\left(\theta /2^{i-1}\right)=2sin\left(\theta /2^i\right)cos\left(\theta /2^i\right)$$$$\frac{cos\left(\theta \right)\cdot cos\left(\theta /2\right)\cdot cos\left(\theta /4\right)\cdot ...\cdot \frac{1}{2}\cdot \frac{1}{2}2cos\left(\theta /2^{i-1}\right)\cdot sin\left(\theta /2^{i-1}\right)}{sin\left(\theta /2^i\right)}$$$$\frac{cos\left(\theta \right)\cdot cos\left(\theta /2\right)\cdot cos\left(\theta /4\right)\cdot ...\cdot \frac{1}{2^2}\cdot \frac{1}{2}2cos\left(\theta /2^{i-2}\right)\cdot sin\left(\theta /2^{i-2}\right)}{sin\left(\theta /2^i\right)}$$$$⋮$$$$\frac{sin\left(2\theta \right)}{2^{i+1}sin\left(\theta /2^i\right)}=\frac{sin\left(2\theta \right)}{2^{i+1}\frac{sin\left(\theta /2^i\right)}{\theta /2^i}\cdot \left(\theta /2^i\right)}$$$$luego$$$$$$$$\underset{i\to \mathrm{\infty }}{\lim }\frac{sin\left(2\theta \right)}{2^{i+1}\frac{sin\left(\theta /2^i\right)}{\theta /2^i}\cdot \left(\theta /2^i\right)}=\frac{sin\left(2\theta \right)}{2^{i+1}\cdot \left(\theta /2^i\right)\underset{i\to \mathrm{\infty }}{\lim }\frac{sin\left(\theta /2^i\right)}{\left(\theta /2^i\right)}}$$$$=\frac{sin\left(2\theta \right)}{2\theta }=\frac{1}{2\theta }\mathrm{ya}\mathrm{que}\theta =\frac{\pi }{4}$$$$\Rightarrow \underset{i=0}{\prod ^{\mathrm{\infty }}}cos\left(\frac{\theta }{2^i}\right)=\frac{1}{\pi /2}=\frac{2}{\pi }$$$$$$$$$$$$$$$$$$

Commented by rahul 19 last updated on 19/Nov/18

thank you so much Arcana ,����

Commented by ajfour last updated on 19/Nov/18

$$superb!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com