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I require some help to push me in the right direction to solve these equations.

$t_1 = P_1\sin(A)\sin(B) + P_2\cos(A)\cos(B)$

$t_2 = P_3\cos(A)\sin(B) + P_4\sin(A)\cos(B)$

where $t_1, t_2, P_1, P_2, P_3, P_4$ are known coefficients. Solving for $A$ and $B$.

Any help greatly appreciated.

This may seem like it, but this is not homework

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    thank you, however i cannot assume P1 = P2 and P3=P42011-07-14

2 Answers 2

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Using the product to sum formulas we can convert your equations into

$ e_1 = a \cos x + b \cos y$

$ e_2 = c \sin x + d \sin y$

where $x = A+B$ and $y = A-B$.

This gives us

$ \left(\frac{a \cos x - e_1}{b}\right)^2 + \left(\frac{c \sin x -e_2}{d}\right)^2 = 1$

which can be made into a 4th degree polynomial in terms of $\cos x$: by moving $\sin x$ term to the right and using $\sin^2 x = 1 - \cos ^2 x$ and squaring, which is solvable in "closed" form and standard techniques exist to do that.

Once you solve for $\cos x$, it is easy to solve for $\cos y$ and thus get back $A$ and $B$.

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Putting $x=\sin A$, $y=\sin B\ $ turns the system into algebraic one: $ \left(t_1-P_1 x y\right){}^2=P_2^2 \left(1-x^2\right) \left(1-y^2\right), $ $ \left(P_3^2 \left(x^2-1\right) y^2+P_4^2 x^2 \left(y^2-1\right)+t_2^2\right){}^2=4 P_3^2 P_4^2 x^2 \left(x^2-1\right) y^2 \left(y^2-1\right). $ For concrete values of parameters its solutions can be obtained in mathematical software.