given-that-is-a-real-number-use-mathematical-induction-or-otherwise-to-show-that-cos-2-cos-2-2-cos-2-3-cos-2-n-sin-2-n-sin-2-n-hence-find-the-lim-n-

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}}) \dots \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}}) \dots \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}}) \dots \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}}) \dots \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}}) \dots 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}}) \dots 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}}) \dots 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}}) \dots \sin (\frac{α}{2^{n - 2}})

\dots \dots

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