I just read somwhere thatif cos a = cos dthen a=2n(pi)±d where

  1. i just read somwhere that

    if cos a = cos d

    then a=2n(pi)±d where n is integer

    So, if cos (b*ln (i/j))= cos (-b (lni/j))=cos(b*ln(j/i))

    can i write
    b*ln i/j = 2n(pi)±b*ln(j/i)
    or
    bln(j/i)+bln(i/j)=2n(pi) or 2n(pi)=0
    b=n(pi)/(ln(i/j)-ln(j/i))=n(pi)/(b*ln(i/j))

    then replace b in
    cos(b*lni/j)=cos(n(pi))= ±1

    I feel that something is wrong. But what?
     
    Last edited: Feb 25, 2012
  2. jcsd
  3. Office_Shredder

    Office_Shredder 4,499
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: Trigonometry

    This isn't entirely true... for example, cos(pi/2) and cos(3pi/2) are equal.

    This is definitely true, but not as interesting as it might look at first sight. Just take n=0 and b*ln(i/j)=-b*ln(j/i) (note that the statement is a=2n(pi)+d for SOME integer n, not every integer)

    When you divide by ln(i/j)+ln(j/i) you divided by zero. I'm unsure where your last equality comes from
     
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