Prove that for every z the product $\prod_{k=1}^{\infty}cos(\frac{z}{2^k}) = \frac{sin(z)}{z}$
[Hint: Use the fact that $sin(2z) = 2*sin(z)*cos(z)$.]
Then my idea was the following, I have that:
$P_n = \prod_{k=1}^{n}cos(\frac{z}{2^k}) = cos(\frac{z}{2}) * cos(\frac{z}{4}) *cos(\frac{z}{8}) *…*cos(\frac{z}{2^n})$
When I use the hint for each one of the cosine terms I have that any terms vanishing then finally
$P_n = \prod_{k=1}^{n}cos(\frac{z}{2^k}) = \frac{sin(z)}{2^n*sin(\frac{z}{2^n})}$
But I can take the $lim_{n→\infty}$ to see if $\prod_{k=1}^{\infty}cos(\frac{z}{2^k}) = \frac{sin(z)}{z}$
Does anyone have an idea, please.
Best Answer
Use the fact that $\frac {\sin (\frac z {2^{n}})} {(\frac z {2^{n}})} \to 1$ as $n \to \infty$.