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Sin product

  1. Dec 26, 2004 #1
    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
    [tex] \sin{x} = 2^n \cdot \sin{\left(\frac{x}{2^n}\right)} \prod_{k=1}^{n} \cos{\left(\frac{x}{2^k}\right)} [/tex]

    Valid for [tex] n \in \mathbb{N} \backslash \{ 0 \} [/tex]

    Can you use this formula for anything?
  2. jcsd
  3. Dec 26, 2004 #2


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    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.
  4. Dec 26, 2004 #3


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

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

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


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