# limiting value of Cosine

#### 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 $$\displaystyle n \to \infty$$.

Last edited by a moderator:

#### Also sprach Zarathustra

Hint:

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

$$\displaystyle sin2x=2sinxcosx$$

#### 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.

#### mr fantastic

MHF Hall of Fame
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:

#### Chandru1

thanks

Mr. Fantastic thanks for the fantastic answer.