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I have the following equation with some trigonometric functions: ( where: $u_{x}, u_{y}, a, b$ are known. To be found are: $\theta,\phi$)

$$ \begin{cases} u_{x} = a\sin(\theta)\cos(\phi) + b\sin(\phi) \\ u_{y} = b\sin(\theta)\cos(\phi) - a\sin(\phi) \end{cases} $$

It seems at first simple but i cannot find the solution. What I did is the following:

  1. I squared both equations and I got: $$ \begin{cases} u_{x}^{2} = ( a\sin(\theta)\cos(\phi) + b\sin(\phi))^{2} \\ u_{y}^{2} = ( b\sin(\theta)\cos(\phi) - a\sin(\phi) )^{2} \end{cases} $$

  2. adding both equations toghether and then simpifing: $$ u_{x}^{2} + u_{y}^{2} = a^{2}\sin(\theta)^{2}\cos(\phi)^{2} + b^{2}\sin(\phi)^{2} + b^{2}\sin(\theta)^{2}\cos(\phi)^{2} + a^{2}\sin(\phi)^{2} $$

  3. I get the following equation:

$$ \frac{u_{x}^{2} + u_{y}^{2}}{a^{2} + b^{2}} = \sin(\theta)^{2}\cos(\phi)^{2} + \sin(\phi)^{2} $$

but I cannot get the final solution. Could you please help me to solve this problem? Regards and thanks

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rewrite like follows
$u_x-b\sin(\phi)=a\sin(\theta)\cos(\phi)$ (1)
$u_y+a\sin(\phi)=b\sin(\theta)\cos(\phi)$ (2) and divide (1) by (2)

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  • $\begingroup$ That's work!!!! Thanks a lot Dr.!!! $\endgroup$
    – Dave
    Commented Oct 28, 2014 at 21:02

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