limiting value of Cosine

Feb 2009
148
10
Chennai
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 \(\displaystyle n \to \infty\).
 
Last edited by a moderator:
Dec 2009
1,506
434
Russia
Hint:

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

\(\displaystyle sin2x=2sinxcosx\)
 
Feb 2009
148
10
Chennai
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.
 

mr fantastic

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Dec 2007
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6,768
Zeitgeist
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.
 
Last edited:
Feb 2009
148
10
Chennai
thanks

Mr. Fantastic thanks for the fantastic answer.