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


by sparsh12
Tags: a2npid, dthen, somwhere, thatif
sparsh12
sparsh12 is offline
#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?
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Office_Shredder
Office_Shredder is offline
#2
Feb25-12, 08:59 PM
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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


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