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Question Number 156891 by cortano last updated on 16/Oct/21

Answered by mindispower last updated on 17/Oct/21

$$1-{tg}^2\left(\frac{x}{2^k}\right)=\frac{cos\left(\frac{x}{2^{k-1}}\right)}{{cos}^2\left(\frac{x}{2^k}\right)}$$$$\underset{k=1}{\prod ^n}(1-{tg}^2\left(\frac{x}{2^k}\right))=\frac{1}{cos\left(\frac{x}{2^n}\right)\underset{k=1}{\prod ^n}cos\left(\frac{x}{2^k}\right)}.\frac{sin\left(\frac{x}{2^n}\right)}{sin\left(\frac{x}{2^n}\right)}$$$$=2^n\frac{sin\left(\frac{x}{2^n}\right)}{cos\left(\frac{x}{2^n}\right)sin\left(x\right)}$$$$\underset{n\to \mathrm{\infty }}{\lim }.\frac{2^{n}sin\left(\frac{x}{2^n}\right)}{cos\left(\frac{x}{2^n}\right)sin\left(x\right)}=\frac{x}{sin\left(x\right)}$$$$\underset{x\to 0}{\lim }\frac{x}{sin\left(x\right)}=\underset{x\to 0}{\lim }\frac{1}{cos\left(x\right)}=1.\left(UsingHopital\right)$$

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