Hi--
Please help me as to what value does $\displaystyle \cos(\theta) \times \cos(\theta/2) \times \cdots \times \cos(\theta/2^{n})$ approach as $\displaystyle n \to \infty$.
Hi--
Please help me as to what value does $\displaystyle \cos(\theta) \times \cos(\theta/2) \times \cdots \times \cos(\theta/2^{n})$ approach as $\displaystyle n \to \infty$.
This infinite product is based on a famous infinite product first discovered by Euler:
$\displaystyle \sin (x) = 2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)$
$\displaystyle = 2^2 \sin \left(\frac{x}{4}\right) \cos \left(\frac{x}{4}\right) \cos \left(\frac{x}{2}\right)$
$\displaystyle = 2^3 \sin \left(\frac{x}{8}\right) \cos \left(\frac{x}{8}\right) \cos \left(\frac{x}{4}\right) \cos \left(\frac{x}{2}\right)$
= ....
$\displaystyle = 2^n \sin \left(\frac{x}{2^n}\right) \cos \left(\frac{x}{2^n}\right) .... \cos \left(\frac{x}{2}\right)$
$\displaystyle = x \cdot \left[ \frac{ \sin \left(\frac{x}{2^n}\right) }{\frac{x}{2^n}} \right] \cos \left(\frac{x}{2}\right) \cos \left(\frac{x}{4}\right).... \cos \left(\frac{x}{2^n}\right)$.
Now take the limit $\displaystyle n \to \infty$:
$\displaystyle \sin (x) = x \cos \left(\frac{x}{2}\right) \cos \left(\frac{x}{4}\right).... \cos \left(\frac{x}{2^n}\right) .... $
I hope you can see what to do with this famous result.