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Question Number 155183 by mathdanisur last updated on 26/Sep/21

Answered by aleks041103 last updated on 26/Sep/21

$$1-{tan}^2\frac{x}{2^k}=\frac{{cos}^2\frac{x}{2^k}-{sin}^2\frac{x}{2^k}}{{cos}^2\frac{x}{2^k}}=\frac{cos\frac{x}{2^{k-1}}}{{cos}^2\frac{x}{2^k}}$$$$\underset{k=1}{\prod ^n}cos\frac{x}{2^{k-1}}=cosxcos\frac{x}{2}...cos\frac{x}{2^{n-1}}=$$$$=\frac{1}{sin\frac{x}{2^{n-1}}}sin\frac{x}{2^{n-1}}cos\frac{x}{2^{n-1}}cos\frac{x}{2^{n-2}}...cos\frac{x}{2}cosx=$$$$=\frac{1}{sin\frac{x}{2^{n-1}}}\frac{1}{2}sin\frac{x}{2^{n-2}}cos\frac{x}{2^{n-2}}...cos\frac{x}{2}cosx=$$$$=\frac{1}{2^{2}sin\frac{x}{2^{n-1}}}sin\frac{x}{2^{n-3}}\underset{k=0}{\prod ^{n-3}}cos\frac{x}{2^k}=...=$$$$=\frac{1}{2^{m}sin\frac{x}{2^{n-1}}}sin\frac{x}{2^{n-1-m}}\underset{k=0}{\prod ^{n-1-m}}cos\frac{x}{2^k}=$$$$=\frac{1}{2^{n-1}sin\frac{x}{2^{n-1}}}sinxcosx=\frac{sin\left(2x\right)}{2^{n}sin\left(\frac{x}{2^{n-1}}\right)}$$$$\underset{k=1}{\prod ^n}cos\frac{x}{2^k}=\underset{k=1}{\prod ^n}cos\frac{x/2}{2^{k-1}}=$$$$=\frac{sin\left(2\left(x/2\right)\right)}{2^{n}sin\left(\frac{x/2}{2^{n-1}}\right)}=\frac{sinx}{2^{n}sin\frac{x}{2^n}}$$$$\Rightarrow \underset{k=1}{\prod ^n}\frac{cos\frac{x}{2^{k-1}}}{{cos}^2\frac{x}{2^k}}=\frac{\left(\underset{k=1}{\prod ^n}cos\frac{x}{2^{k-1}}\right)}{{\left(\underset{k=1}{\prod ^n}cos\frac{x}{2^k}\right)}^2}=$$$$=\frac{\frac{sin\left(2x\right)}{2^{n}sin\left(\frac{x}{2^{n-1}}\right)}}{{\left(\frac{sinx}{2^{n}sin\frac{x}{2^n}}\right)}^2}=\frac{sin\left(2x\right)2^{2n}{sin}^2\frac{x}{2^n}}{2^{n}sin\frac{2x}{2^n}{sin}^{2}x}=$$$$=\frac{2sinxcosx{2}^n{sin}^2\frac{x}{2^n}}{2sin\frac{x}{2^n}cos\frac{x}{2^n}{sin}^{2}x}=\frac{2^{n}cosxsin\frac{x}{2^n}}{cos\frac{x}{2^n}sinx}=$$$$=2^n\frac{tan\left(\frac{x}{2^n}\right)}{tan\left(x\right)}$$$$\Rightarrow \mathrm{\Omega }=\underset{x\to 0}{lim}\frac{x}{tan\left(x\right)}\underset{n\to \mathrm{\infty }}{lim}\frac{tan\left(x/2^n\right)}{x/2^n}=1$$

Commented by aleks041103 last updated on 26/Sep/21

$$Thereisaneveneasiermethod$$$$tan\left(2x\right)=\frac{2tan\left(x\right)}{1-{tan}^2\left(x\right)}\Rightarrow 1-{tan}^{2}x=2\frac{tan\left(x\right)}{tan\left(2x\right)}$$$$\Rightarrow \underset{k=1}{\prod ^n}(1-{tan}^2\frac{x}{2^k})=2^n\underset{k=1}{\prod ^n}\frac{tan\left(\frac{x}{2^k}\right)}{tan\left(\frac{x}{2^{k-1}}\right)}$$$$telescopicsum:$$$$\Rightarrow \underset{k=1}{\prod ^n}(1-{tan}^2\frac{x}{2^k})=\frac{2^{n}tan\left(2^{-n}x\right)}{tan\left(x\right)}$$$$...$$$$\Rightarrow \mathrm{\Omega }=1$$

Commented by mathdanisur last updated on 26/Sep/21

$$\mathrm{Very}\mathrm{nice}\mathbf{S}\mathrm{er},\mathrm{thank}\mathrm{you}$$

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