folks! I am getting the following system of trigonometric equations which has come as a product of my research. But, I have tried a lot with no success, to solve this system. Can anyone help? Thank you.

Find $\theta_1$ and $\theta_2$.

$$\left(\frac{\cos\alpha_1}{\cos\theta_1}\right)^2+\left(\frac{\cos\alpha_2}{\cos\theta_2}\right)^2=A\quad(A \ \mbox{is a constant},\ A\in(0,1))\tag{1}$$ $$\frac{1}{\cos^2\alpha_1}\left(\frac{1}{\sin^4\theta_1}-1\right)=\frac{1}{\cos^2\alpha_2}\left(\frac{1}{\sin^4\theta_2}-1\right)\tag{2} $$

Note: I know that I can reduce this to finding roots of a polynomial, but I wonder if there is an elegant method to find a solution to this problem.

Edit: The equations @DavidHolden has written are closer to the actual equations I wanted to solve. The algebraic version of the equations that I want to solve is the following

Find $x_1,x_2\quad 0\le a_1\le x_1\le 1, 0\le a_2\le x_2\le 1$ such that

$$x_1+x_2=A\quad \tag{1}$$ $$a_1\left(\frac{1}{x_1^2}-\frac{1}{(x_1-a_1)^2}\right)=a_2\left(\frac{1}{x_2^2}-\frac{1}{(x_2-a_2)^2}\right)\tag{2}$$ Note that if in the Equation(2) instead of $a_1,\ a_2$, we have $a_1^2,\ a_2^2$ multiplied then we immediately have the solution $$x_1=\frac{a_1}{a_1+a_2}A,\ x_2=\frac{a_2}{a_1+a_2}A$$

Further Edit: The Problem can be restated as below: With $0\le a_1\le x_1\le 1,\ 0\le a_2\le x_2\le 1$ find $x_1,\ x_2$ such that $x_1+x_2=A$ and $$f'(x_1,a_1)=f'(x_2,a_2)$$ where $$f(x,a)=g(x/a);\quad g(x)=\frac{1}{x(x-1)}$$


1 Answer 1


although not much of a step in solving your problem, perhaps it is of use to simplify a little, so certain salient features are more clearly visible. if we set $x_i = cos^2 \; \theta_i$, and $a_i=cos^2 \alpha_i $ then the equations become: $$\frac{a_1}{x_1} +\frac{a_2}{x_2} =A \\ \frac{a_2}{(1-x_1)^2} - \frac{a_1}{(1-x_2)^2} = a_2-a_1 $$ is this the sort of thing you meant when you mentioned reducing this to finding roots of a polynomial?

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    $\begingroup$ Minor suggestion: If you re-combine the $a_i$ terms, and factor, then you can write the second equation as $$\frac{a_1}{x_1} \frac{2-x_2}{(1-x_2)^2} = \frac{a_2}{x_2}\frac{2-x_1}{(1-x_1)^2}$$ This might be considered slightly nicer representation, in that the $a_i/x_i$ fractions appear in both equations. (It doesn't make finding a solution any easier. Eliminating one of the $x_i$s leaves a messy cubic in the other.) $\endgroup$
    – Blue
    Commented Dec 22, 2013 at 13:48
  • $\begingroup$ very nice, @Blue. this shows that we may define a function: $f(x)=\frac{x(2-x)}{(1-x)(1-x)}$, in terms of which: $\frac{f(x_1)}{a_1} = \frac{f(x_2)}{a_2}$ $\endgroup$ Commented Dec 22, 2013 at 13:52
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    $\begingroup$ @DavidHolden actually the equations you have written are the ones which I got initially :-). To solve them I used the transformations $x_1=\cos^2\theta_1,\ x_2=\cos^2\theta_2,\ a_1=\cos^2\alpha_1,\ a_1=\cos^2\alpha_2$ to get the equations in the question. Anyway, the equations that you have got are the ones I want to solve. $\endgroup$ Commented Dec 22, 2013 at 14:11
  • $\begingroup$ @SamratMukhopadhyay ;-) a rather nice little irony. sorry to retrace your steps... $\endgroup$ Commented Dec 22, 2013 at 14:19
  • $\begingroup$ It's ok, i was just trying to see if trigonometry can help me solve this. $\endgroup$ Commented Dec 22, 2013 at 14:22

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