Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A problem about sin and cos

  1. Feb 19, 2005 #1
    Given that


    n = Π[2 – 2cos(kπ/ n)] ........ (where Π is the product sign , from k = 1 to n-1 )


    as
    cos2@ = 1 – 2(sin@)^2

    then
    2 – 2cos(kπ/ n) = 4[sin(kπ/ 2n)]^2 , for k = 1 , 2 , 3 , … n-1

    then
    n = Π[4[sin(kπ/ 2n)]^2] = [4^(n-1)] Π[sin(kπ/ 2n)]^2

    but the book then said
    Π[sin(kπ/ n)]^2 = n / [4^(n-1)]

    why ?
    why is not the sin(kπ/ 2n) but sin(kπ/ n) ?
     
  2. jcsd
  3. Feb 19, 2005 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    It's definitely a typo.It must be the "2" in the denominator.

    Daniel.
     
  4. Feb 19, 2005 #3
    thank you first .


    actually , the question is asking me to show :

    [tex] $ \prod_{1}^{n-1} \sin{(\frac{k\pi}{n})} = \frac{n^{0.5}}{2^{n-1}}[/tex]

    by finding the roots of [tex] \ x^{2n} - 1 = 0[/tex]

    is the question wrong or I have made misstake?
     
  5. Feb 20, 2005 #4

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Again it's the "2" in the denominator missing...As for the equation,solve it and see whether you can relate the solutions to the identity which you have proven.

    Daniel.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: A problem about sin and cos
  1. Sin,cos problems (Replies: 1)

  2. Sin and Cos (Replies: 7)

  3. Sin/cos/tan by hand? (Replies: 2)

Loading...