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

1. Feb 25, 2012

### sparsh12

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. Feb 25, 2012

### Office_Shredder

Staff Emeritus
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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