How can I solve a problem with two equations in two unknowns using trigonometry?

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The discussion focuses on solving a system of two equations in two unknowns using trigonometric identities. The equations presented are C = Acos(x) + Bcos(y) and 0 = Asin(x) + Bsin(y), where A, B, and C are known constants. The solution involves expressing cos(y) and sin(y) in terms of x, followed by applying the Pythagorean identity to derive a single equation that can be solved for x. This method allows for the subsequent determination of y by substitution.

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

[tex]\begin{array}{l}<br /> \cos (y) = \frac{{C - A\cos (x)}}{B}\\<br /> \sin (y) = - \frac{{A\sin (x)}}{B}<br /> \end{array}[/tex]

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

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

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.
 

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