given that α is a real number, use mathematical induction or otherwise to show that cos ((α/2))cos((α/2^2 ))cos((α/2^3 )) ...cos((α/2^n )) = ((sin α)/(2^n sin((α/2^n )))) hence find the lim_(n→∞) cos((α/2))cos((α/2^2 ))cos((α/2^3 )) ... cos((α/2^n ))


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Question Number 93365 by Rio Michael last updated on 12/May/20

$given that α is a real number, use mathematical induction or$ $otherwise to show that$ $\cos (\frac{α}{2}) \cos (\frac{α}{2^{2}}) \cos (\frac{α}{2^{3}}) . . . \cos (\frac{α}{2^{n}}) = \frac{\sin α}{2^{n} \sin (\frac{α}{2^{n}})}$ $hence find the$ $\lim_{n \to \infty} \cos (\frac{α}{2}) \cos (\frac{α}{2^{2}}) \cos (\frac{α}{2^{3}}) . . . \cos (\frac{α}{2^{n}})$
Commented by mr W last updated on 12/May/20

$P = \cos (\frac{α}{2}) \cos (\frac{α}{2^{2}}) \cos (\frac{α}{2^{3}}) . . . \cos (\frac{α}{2^{n}})$ $P 2 \sin (\frac{α}{2^{n}}) = \cos (\frac{α}{2}) \cos (\frac{α}{2^{2}}) \cos (\frac{α}{2^{3}}) . . . \cos (\frac{α}{2^{n}}) 2 \sin (\frac{α}{2^{n}})$ $P 2 \sin (\frac{α}{2^{n}}) = \cos (\frac{α}{2}) \cos (\frac{α}{2^{2}}) \cos (\frac{α}{2^{3}}) . . . co s (\frac{α}{2^{n - 1}}) \sin (\frac{α}{2^{n}})$ $P 2^{2} \sin (\frac{α}{2^{n}}) = \cos (\frac{α}{2}) \cos (\frac{α}{2^{2}}) \cos (\frac{α}{2^{3}}) . . . co s (\frac{α}{2^{n - 1}}) 2 \sin (\frac{α}{2^{n}})$ $P 2^{2} \sin (\frac{α}{2^{n}}) = \cos (\frac{α}{2}) \cos (\frac{α}{2^{2}}) \cos (\frac{α}{2^{3}}) . . . co s (\frac{α}{2^{n - 2}}) \sin (\frac{α}{2^{n - 1}})$ $P 2^{3} \sin (\frac{α}{2^{n}}) = \cos (\frac{α}{2}) \cos (\frac{α}{2^{2}}) \cos (\frac{α}{2^{3}}) . . . \sin (\frac{α}{2^{n - 2}})$ $. . . . . .$ $P 2^{n - 1} \sin (\frac{α}{2^{n}}) = \cos (\frac{α}{2}) \sin (\frac{α}{2})$ $P 2^{n} \sin (\frac{α}{2^{n}}) = \cos (\frac{α}{2}) 2 \sin (\frac{α}{2})$ $P 2^{n} \sin (\frac{α}{2^{n}}) = \sin (α)$ $\Rightarrow P = \frac{\sin (α)}{2^{n} \sin (\frac{α}{2^{n}})}$ $\Rightarrow \lim_{n \to \infty} P = \frac{\sin (α)}{α} \times \frac{(\frac{α}{2^{n}})}{\sin (\frac{α}{2^{n}})} = \frac{\sin (α)}{α}$
Commented by Rio Michael last updated on 12/May/20

$perfect sir$
Commented by Rio Michael last updated on 12/May/20

$Q 93368 sir pleasecan you try this one$
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