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

by sparsh12
Tags: a2npi±d, dthen, somwhere, thatif
 P: 13 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?
Emeritus
PF Gold
P: 4,500
 Quote by sparsh12 i just read somwhere that if cos a = cos d then a=2n(pi)±d where n is integer
This isn't entirely true... for example, cos(pi/2) and cos(3pi/2) are equal.

 b*ln i/j = 2n(pi)±b*ln(j/i)
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)

 b=n(pi)/(ln(i/j)+ln(j/i))=n(pi)/(b*ln(i/j))
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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