Solving 2 Simultaneous Equations for c & d

[ChrisHarvey]

Can anyone analytically solve these 2 simultaneous equations for c & d, where a and b are known?

a = sin(c) * cosh(d)
b = cos(c) * sinh(d)

I can eliminate 'c' to get:

(sinh d)^2 = b^2 + (a^2)*(tanh d)^2

but this doesn't seem to make life any easier.

 

[ChrisHarvey]

Arggghhhh... I've just this second seen how to do it: write (tanh d)^2 as 1/(cosh d)^2 and then (cosh d)^2 as 1 + (sinh d)^2, multiply though by (sinh d)^2, and solve as a quadratic.

I have an annoying habit of asking for help and then doing it myself a few minutes later... sorry.

 

[tacman]

Yes, it is possible to solve these simultaneous equations for c and d analytically. Here is one way to approach it:

1. Rearrange the equations to isolate the sine and cosine terms:
c = arcsin(a/cosh(d))
c = arccos(b/sinh(d))

2. Set the two expressions for c equal to each other:
arcsin(a/cosh(d)) = arccos(b/sinh(d))

3. Use the trigonometric identity sin(x) = cos(π/2 - x) to rewrite the left side:
arcsin(a/cosh(d)) = π/2 - arccos(a/cosh(d))

4. Simplify the right side using the inverse cosine identity:
arcsin(a/cosh(d)) = π/2 - arcsin(b/sinh(d))

5. Set the expressions inside the inverse sine equal to each other and solve for d:
a/cosh(d) = b/sinh(d)
d = arctanh(b/a)

6. Substitute this value for d into either of the original equations to solve for c:
c = arcsin(a/cosh(arctanh(b/a))) or c = arccos(b/sinh(arctanh(b/a)))

This solution may seem complicated, but it is an analytical solution that can be used to find the exact values of c and d for any given values of a and b. It may be helpful to use a calculator or software program to evaluate the inverse trigonometric functions and simplify the expressions.