can u convert to python code these solutions, these python code must run, thanks a lot

where v/A and /A are the velocity and angular velocity of point A in reference frame A, respectively. We can express these velocities in terms of the angular velocities and orientation angles given in the problem statement: v/A=1/2(sin()x+cos()sin()*y+cos()cos()z) /A=sin()x+cos()cos()y cos()sin()cos()z Substituting these equations into the expression for BA, we get: B,A=1/2(sin()xcos()sin()ycos()cos()z)bx+(sin()x+cos()cos()ycos()sin() Next, we need to find GB, the angular velocity of B in reference frame G. From the problem statement, we know that G B is given by: GB=xbx+yby+zbz Finally, we can substitute BA and GB into the equation for GA, and simplify: GA=GB+BA GA=xbx+yby+zbzI/2(sin()xcos()sin()ycos()cos()z)bx+(sin()x+cos()cos()ycos()sin()cos()z)by Therefore, the solution is: \[ \begin{array}{l} \text { G_omega_A }=\omega x * b^{\wedge} x+\omega y * b^{\wedge} y+\omega z * \\ b^{\wedge} z-l / 2 *(\sin (\theta) * \omega x-\cos (\theta) * \sin (\phi) * \\ \omega y-\cos (\theta) * \cos (\phi) * \omega z) b^{\wedge} x+(\sin (\beta) * \\ \omega x+\cos (\beta) * \cos (\theta) * \omega y-\cos (\beta) * \sin (\theta) \\ * \cos (\phi) * \omega z) b^{\wedge} y \end{array} \]