- #1
sparsh12
- 12
- 0
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?
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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