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.