Re: Reducing (1-a^2)*(1-b^2)=(1-c^2)*(1-ab)^2
- From: José Carlos Santos <jcsantos@xxxxxxxx>
- Date: Fri, 24 Aug 2007 15:37:22 +0100
On 24-08-2007 15:21, Peter Christensen wrote:
Are there anybody with a good idea?
I would like to isolate a or b.
I don't understand your question. Are you asking how to get _a_ from
_b_? That is easy:
a = (c - b)/(b c - 1) or a = (c + b)/(b c + 1)
After all, it amounts to solving a second degree equation.
Best regards,
Jose Carlos Santos
.
- Follow-Ups:
- Re: Reducing (1-a^2)*(1-b^2)=(1-c^2)*(1-ab)^2
- From: Peter Christensen
- Re: Reducing (1-a^2)*(1-b^2)=(1-c^2)*(1-ab)^2
- References:
- Reducing (1-a^2)*(1-b^2)=(1-c^2)*(1-ab)^2
- From: Peter Christensen
- Reducing (1-a^2)*(1-b^2)=(1-c^2)*(1-ab)^2
- Prev by Date: Re: Math is not a memoriter course.
- Next by Date: Re: Reducing (1-a^2)*(1-b^2)=(1-c^2)*(1-ab)^2
- Previous by thread: Reducing (1-a^2)*(1-b^2)=(1-c^2)*(1-ab)^2
- Next by thread: Re: Reducing (1-a^2)*(1-b^2)=(1-c^2)*(1-ab)^2
- Index(es):
Relevant Pages
|
