a*sin(theta)+b       a*cos(theta) + e
--------------  =   ----------------
c*sin(theta)+d       c*cos(theta) + f

solved for theta by Scott Datallo's magic gives:

theta = arcsin(-C * cos(gamma)) - gamma

where C = (de-bf)/(ec-af)
and    gamma = arctan( (ad-bc)/(ec-af) )


Scott, you nailed it!

It took me a bit to figure out a couple of your steps, but then I remembered 
my sum-of-angles formulae and off I went.

Did you do this by hand or use some tool? What prompted the brilliant 
substitution of tan(gamma) = (ad-bc)/(ec-af) ??

I am in awe!

Bob Ammerman
RAm Systems


----- Original Message ----- 
From: "Scott Dattalo" <scott@dattalo.com>
To: "Microcontroller discussion list - Public." <piclist@mit.edu>
Sent: Friday, April 01, 2005 4:27 PM
Subject: Re: [EE] Math formula help.


> Bob Ammerman wrote:
>
>>> you will note of course this is the same as:
>>>
>>>
>>> where x = sin(theta)  (and 'theta' is indeed an angle in the range 
>>> [0..2pi) radians
>
> Bob,
>
> If you cross multiply and collect terms you get:
>
>   tan(gamma) * cos(theta) +  sin(theta) + C = 0
>
> Where
>
>  tan(gamma) = (ad-bc)/(ec-af)
>  C = (de-bf)/(ec-af)
>
> This can be simplified to:
>
>   sin(theta + gamma) = -C * cos(gamma)
>
> and solved for theta:
>
>   theta = arcsin(-C * cos(gamma)) - gamma
>
> of course,
>
>   gamma = arctan( (ad-bc)/(ec-af) )
>
> Scott
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