Register to reply

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

by sparsh12
Tags: a2npid, dthen, somwhere, thatif
Share this thread:
sparsh12
#1
Feb25-12, 08:00 PM
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?
Phys.Org News Partner Mathematics news on Phys.org
Researcher figures out how sharks manage to act like math geniuses
Math journal puts Rauzy fractcal image on the cover
Heat distributions help researchers to understand curved space
Office_Shredder
#2
Feb25-12, 08:59 PM
Emeritus
Sci Advisor
PF Gold
P: 4,500
Quote Quote by sparsh12 View Post
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


Register to reply

Related Discussions
Hi, How do I find the height of the triangle , the triangle is set Precalculus Mathematics Homework 9
Trigonometry help General Math 7
Find the period of the following function Calculus & Beyond Homework 1