Re: Simple trig question
- From: "jmorriss@xxxxxxxxxxx" <jmorriss@xxxxxxxxxxx>
- Date: Mon, 09 Jul 2007 12:32:02 -0700
On Jul 9, 3:13 pm, Eric C <eric.fo...@xxxxxxxxx> wrote:
I am currently learning more about SAR radar processing and I'm
reading a Spotlight Mode SAR book by Jakowatz. I enjoy the book, but
I've hit a very minor stumbling block I can't seem to figure out. It's
on page 361, for those who have the book, but I'll simplify it here.
It involves a triangle with sides a, r0, and r, where a << r0. He
first sites the law of cosines:
r^2 = r0^2 - a*a*r0*cos(phi)
where phi is the angle across from r0. Then he states at a << r0 (also
r0 > r), therefore the following is approximately true:
r - r0 = -a cos(phi) + (a^2 / (2*r0) ) * sin(phi)^2
For the life of me I cannot figure out where that second term is
coming from. I know it's basic, I just can't find it. I've tried
looking at the problem from multiple angles, so to speak, but I just
can't see that term. I've come up with approximations of my own. Can
anyone explain that to me?
By the way, he also says (also approximately equals)
(r - r0)^2 = a^2 * cos(phi)^2
Which makes sense to me.
Well, for starters, the version of the Law of Cosines as printed in
your post is incorrect (IMHO).
If the sides are r, ro, and a, and the angle given is phi, opposite
the angle opposite ro, then the LofC says:
ro^2 = a^2 +r^2 - 2 * a * r*cos( phi)
Rearranging to give the LHS of the form you quote:
r^2 = ro^2 - a^2 + 2 * a * r*cos( phi)
which is not what you quote for the RHS...
.
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