# A little bit of trigonometry

1. Sep 18, 2010

### cap'n ahab

Hi, I have a problem that arose in one of my courses that I've gotten stuck on. I reduced it to two equations in two unknowns but can't get any further. I've apparently forgotten all the trigonometry I used to know.

C = Acos(x) + Bcos(y)

0 = Asin(x) + Bsin(y)

where A,B,C are known but tedious to write (they are on the order of 10^(-19)).

I'd appreciate any help you can offer. I'm sure the solution will be obvious once I see it.

2. Sep 19, 2010

### danago

First thing that comes to my mind is to first express the equations in the following form:

$$\begin{array}{l} \cos (y) = \frac{{C - A\cos (x)}}{B}\\ \sin (y) = - \frac{{A\sin (x)}}{B} \end{array}$$

You can then use the pythagorean identity ($${\sin ^2}(y) + {\cos ^2}(y) \equiv 1$$) to form the following equation:

$${\left( {\frac{{A\sin (x)}}{B}} \right)^2} + {\left( {\frac{{C - A\cos (x)}}{B}} \right)^2} = 1$$

Expand it out and it should be easy to solve for x, which can then be substituted into one of the first equations to find y.