Continued Cosine Product.

Question

Is there a way to evaluate,

$$ \large \cos x \cdot \cos \frac{x}{2} \cdot \cos \frac{x}{4} ... \cdot \cos \frac{x}{2^{n-1}} \tag*{(1)} $$

I asked this to one of my teachers and what he told is something like this,

Multiply and divide the last term of $(1)$ with $\boxed{\sin \frac{x}{2^{n-1}}}$

So, $$ \large \frac{\cos \frac{x}{2^{n-1}} \cdot \sin \frac{x}{2^{n-1}}}{\sin \frac{x}{2^{n-1}}} \\ \tag*{(2)} $$ $$ \large \implies \frac{\sin (2x)}{2^n \cdot \sin \frac{x}{2^{n-1}}} \\ $$ $$ \large \implies \frac{\sin (2x)}{2^n \cdot \frac{\sin \frac{x}{2^{n-1}}}{\frac{x}{2^{n-1}}} \cdot \frac{x}{2^{n-1}}} \tag*{(3)} $$

Now, as $n \to \infty$ , we have $x \to 0$, Using this, $\lim$ we have,

$$ \boxed{ \lim_{x \to 0} \frac{\sin x}{x} = 1} $$

Using this in $(3)$, we have, $$ \large \boxed{\frac{\sin (2x)}{2x}} \tag*{(4)} $$

All the steps sort of make sense. My doubts are,

How do I do this for other trigonometric ratios?
How does the step 2 happen?

I need help looking into it more intuitionally.

Please provide necessary reading suggestions.

Regards.

I think you mean $x/2^{n-1}\to 0$ as $n\to\infty$. You have a fixed $x$, and you need to be careful about terminology. — Thomas Andrews, May 03 '17 at 21:16

score 8 · Accepted Answer · answered May 03 '17 at 21:26

By the sine duplication formula $\sin(2z)=2\sin(z)\cos(z)$ we have $\cos(z)=\frac{1}{2}\cdot\frac{\sin(2z)}{\sin(z)}$.
In particular $$ \prod_{k=0}^{n-1}\cos\left(\frac{x}{2^k}\right)=\frac{1}{2^n}\prod_{k=0}^{n-1}\frac{\sin\frac{x}{2^{k-1}}}{\sin\frac{x}{2^k}}=\frac{\sin(2x)}{2^n\sin\frac{x}{2^{n-1}}} \tag{1} $$ is a simple telescopic product, that can be written in the form $$ \frac{\sin(2x)}{2x}\cdot\frac{x}{2^{n-1}\sin\frac{x}{2^{n-1}}}\tag{2}$$ and since $\lim_{w\to 0}\frac{\sin w}{w}=\lim_{w\to 0}\frac{w}{\sin w}=1$ it follows that $$ \prod_{k\geq 0}\cos\left(\frac{x}{2^k}\right)=\frac{\sin(2x)}{2x}.\tag{3}$$ In general, as soon as we have a differentiable function $f(x)$ such that $f(2x)=2 g(x)\,f(x)$ and $f(0)=0, f'(0)\neq 0$ we have the identity $$ \prod_{k\geq 0}g\left(\frac{x}{2^k}\right) = \frac{f(2x)}{2x}\cdot \frac{1}{f'(0)}\tag{4} $$ by de l'Hospital theorem.

How do I do this for continued $\sin$ or $\tan$? Similar approach? — Pragyaditya Das, May 04 '17 at 04:44
@PragyadityaDas: $\prod_{k=0}^{n}\sin\left(\frac{x}{2^k}\right)$ converges to zero really fast since $x\approx 0$ implies $\sin(x)\approx x$. The same applies by replacing $\sin$ with $\tan$, hence the only interesting product is associated with the cosine function. — Jack D'Aurizio, May 04 '17 at 07:54

score 1 · Answer 2 · edited Jan 02 '21 at 18:52

How does step 2 happen:

multiplying the last factor by $\frac {\sin \frac{x}{2^{n-1}}}{\sin \frac{x}{2^{n-1}}}$

gives us

$\cos x \cdot \cos \frac x2\cdots \cos \frac x{2^{n-2}}\cdot \frac{\cos \frac{x}{2^{n-1}} \cdot \sin \frac{x}{2^{n-1}}}{\sin \frac{x}{2^{n-1}}}$

Double angle formula.

$\cos \frac{x}{2^{n-1}} \cdot \sin \frac{x}{2^{n-1}} = \frac 12 \sin \frac{x}{2^{n-2}}$

Applying this we get:

$\cos x \cdot \cos \frac x2\cdots \cdot \frac{\cos \frac x{2^{n-2}}\cdot\sin \frac{x}{2^{n-2}}}{2\sin \frac{x}{2^{n-1}}}$

And we can apply the double angle formula again. And do it repeatedly until all of the $\cos$ factors have been devoured.

Continued Cosine Product.

2 Answers2