# limiting value of Cosine

• Jun 27th 2010, 06:29 PM
Chandru1
limiting value of Cosine
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 $\displaystyle n \to \infty$.
• Jun 27th 2010, 06:41 PM
Also sprach Zarathustra
Hint:

Multiply (and divide) by $\displaystyle 2sin\frac{\theta}{2}$ and use:

$\displaystyle sin2x=2sinxcosx$
• Jun 27th 2010, 07:41 PM
Chandru1
I was solving an analysis problem and i need this quantity to have 1/2^{n} multiplied by some value. If this doesn't work out then my claim would be incorrect.
• Jun 28th 2010, 01:24 AM
mr fantastic
Quote:

Originally Posted by Chandru1
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.
• Jun 28th 2010, 02:45 AM
Chandru1
thanks
Mr. Fantastic thanks for the fantastic answer.