Find u as function of x and y using Python Sympy module when u is given in 2 equations-CodePudding

My goal is to find u (and v) as function of x and y using Sympy module. the equations are:

the answer should be: Ans

from sympy import cosh, sinh, symbols, sin, cos, Eq
u, v, x, y = symbols('u, v, x, y')
eqq1 = Eq(x, sin(u)*cosh(v))
eqq2 = Eq(y, cos(u)*sinh(v))

what's next?

I tried

result = solve((Eqq1, Eqq2), u, v)

obviously its not the right way

CodePudding user response：

Probably not the answer that you expect, but you can rework the problem by eliminating v, as follows:

(x cos u)² - (y sin u)² = cos²u sin²u

Then with t = sin²u, the equation is quadratic:

x² (1 - t) - y² t = (1 - t) t

CodePudding user response：

You already know u as a function of x and y so there remains only to find v in terms of them.

Divide the sides of the equation for y by the corresponding sides of the equation for x and solve for v:

>>> V = solve(Eq(y/x,cos(u)/sin(u)*sinh(v)/cosh(v)),v)
>>> a = sqrt((x*cos(u)   y*sin(u))/(x*cos(u) - y*sin(u)))
>>> assert V == [log(-a), log(a)]

So for each value of u (whose value you already gave as a function of x and y) there are two values of v.

>>> U = asin((sqrt(y**2   (x   1)**2) - sqrt(y**2   (x - 1)**2)/2)
>>> xy={x:.2,y:.1}
>>> ui = U.xreplace(xy).n(3)
>>> vi = [i.xreplace(xy).xreplace({u:ui}).n(3) for i in V];(ui,vi)
(0.929, [0.809   3.14*I, 0.809])