Sin product

[hedlund]

Starting with:

sin(x) = 2sin(x/2)cos(x/2)
sin(x/2) = 2sin(x/4)cos(x/4)
sin(x/4) = 2sin(x/8)cos(x/8) ...

So we can arrive at this
\sin{x} = 2^n \cdot \sin{\left(\frac{x}{2^n}\right)} \prod_{k=1}^{n} \cos{\left(\frac{x}{2^k}\right)}

Valid for n \in \mathbb{N} \backslash \{ 0 \}

Can you use this formula for anything?

[Hurkyl]

Works for n = 0 too. (The empty product is usually defined to be 1)

I've seen it used, I think, the other way around -- to convert a product of cosines into something simpler.

[dextercioby]

Let's check his formula.
\sin x=2\sin(\frac{x}{2})\cos(\frac{x}{2})
\sin(\frac{x}{2})=2\sin(\frac{x}{2^{2}})\cos(\frac {x}{2^{2}})
\sin(\frac {x}{2^{2}})=2\sin(\frac{x}{2^{3}})\cos(\frac{x}{2^ {3}})
...
\sin(\frac{x}{2^{n-1}})=2\sin(\frac{x}{2^{n}})\cos(\frac{x}{2^{n}})

This is a set of "n" equalities.Multiply all relations,simplify through identical terms and everything comes out to be
\sin x= 2^{n}\sin(\frac{x}{2^{n}})\prod_{k=0}^{n}\cos(\fra c{x}{2^{k}})

Just checkin'... :tongue2: Never seen it in my life...

Daniel.

PS.Pretty useful iff u get from somewhere a product of cosine's with argumeents decaying exponentially at a rate of '2'.