Question:

# How to analytically solve this equation with trigonometric function

Abigail: 26 June 2022

The equation containing trigonometric function is:

``````R == Sqrt[(i1-Sin[t])^2 + j1^2 + (k1-Cos[t])^2]*Sqrt[(i2-Sin[t])^2 + j2^2 + (k2-Cos[t])^2]*c/Sqrt[(Sqrt[(i1-Sin[t])^2 + j1^2 + (k1-Cos[t])^2]+Sqrt[(i2-Sin[t])^2 + j2^2 + (k2-Cos[t])^2]+c)*(-Sqrt[(i1-Sin[t])^2 + j1^2 + (k1-Cos[t])^2]+Sqrt[(i2-Sin[t])^2 + j2^2 + (k2-Cos[t])^2]+c)*(Sqrt[(i1-Sin[t])^2 + j1^2 + (k1-Cos[t])^2]-Sqrt[(i2-Sin[t])^2 + j2^2 + (k2-Cos[t])^2]+c)*(Sqrt[(i1-Sin[t])^2 + j1^2 + (k1-Cos[t])^2]+Sqrt[(i2-Sin[t])^2 + j2^2 + (k2-Cos[t])^2]-c)]
``````

The t falls in range of `(0,Pi/2)`. I want to solve `t`. How to make it?

This equation is too complicated and my computer could solve it for a long time?

Easton: 26 June 2022

``````eq1 = R == Sqrt[(i1 - Sin[t])^2 + j1^2 + (k1 - Cos[t])^2]*
Sqrt[(i2 - Sin[t])^2 + j2^2 + (k2 - Cos[t])^2]*
c/Sqrt[(Sqrt[(i1 - Sin[t])^2 + j1^2 + (k1 - Cos[t])^2] +
Sqrt[(i2 - Sin[t])^2 + j2^2 + (k2 - Cos[t])^2] +
c)*(-Sqrt[(i1 - Sin[t])^2 + j1^2 + (k1 - Cos[t])^2] +
Sqrt[(i2 - Sin[t])^2 + j2^2 + (k2 - Cos[t])^2] +
c)*(Sqrt[(i1 - Sin[t])^2 + j1^2 + (k1 - Cos[t])^2] -
Sqrt[(i2 - Sin[t])^2 + j2^2 + (k2 - Cos[t])^2] +
c)*(Sqrt[(i1 - Sin[t])^2 + j1^2 + (k1 - Cos[t])^2] +
Sqrt[(i2 - Sin[t])^2 + j2^2 + (k2 - Cos[t])^2] - c)];
``````

Let us transform it as follows:

``````eq2 = eq1 /. Sin[t] -> Sqrt[1 - Cos[t]^2] /. Cos[t] -> x
``````

That gives a rather long expression. For this reason I do not write it here. It is in terms of `x=Cos[t]`. If we get `x`, we can then calculate arccos(x), provided `-1<x<1`. OK, let us solve it:

``````Solve[eq2, x]
``````

It is solved exactly in somewhat about 10 min, but the result is a huge analytic expression, such that one can hardly make use of it.

Depending on the nature of the problem in such a situation I would look for some sort of a simplification of this equation.

Have fun!