Invert Trigonometric System of Equations

Question

In my research, I have a system of trig equations that I would like to invert. I know $\alpha$ and $\beta$ ($\theta$ is a constant) in the following system.

$$\cos \alpha = \cos \theta \cos x + \sin \theta \sin x \cos y$$ $$\tan \beta = \frac{\cos x +\cos \theta \cos \alpha}{\sin \theta \sin x \sin y} $$

I would like to find an inversion in the form: $$\text{trig}(x) =\; ...$$ $$\text{trig}(y) = \;...$$

How can this be done?

Answer 1

$$\cos(\alpha) = \cos (\theta) \cos( x) + \sin (\theta) \sin (x) \cos( y)\tag 1$$ $$\tan (\beta) = \frac{\cos (x) +\cos (\theta) \cos (\alpha)}{\sin (\theta) \sin (x) \sin (y)}\tag 2$$

From $(1)$ $$\cos(y)=\csc (\theta ) \csc (x) (\cos (\alpha )-\cos (\theta ) \cos (x))\tag 3$$ From $(2)$ $$\sin(y)=\cos (\alpha ) \cot (\beta ) \cot (\theta ) \csc (x)+\cot (\beta ) \csc (\theta ) \cot (x)\tag 4$$ Using $\cos^2(y)+\sin^2(y)=1$, we end with $$1=(\cos (\alpha ) \cot (\beta ) \cot (\theta ) \csc (x)+\cot (\beta ) \csc (\theta ) \cot (x))^2+(\cot (\theta ) \cot (x)-\cos (\alpha ) \csc (\theta ) \csc (x))^2$$

Now, using the tangent half-angle substitution $x=2 \tan ^{-1}(t)$, we end with a quadratic in $t^2$ $$a + b t^2+c t^4 =0$$ where $$a=16 \cos (\alpha ) \cos (2 \beta ) \cot (\theta ) \csc (\theta )+4 \sin ^2(\alpha ) \cos (2 \beta )+$$ $$\csc ^2(\theta ) (\cos (2 \alpha ) (\cos (2 \theta )+3)+9)+3 \cot ^2(\theta )-3$$ $$b=-2 \cos (2 (\alpha -\beta ))-2 \cos (2 (\alpha +\beta ))-16 \sin ^2(\alpha ) \csc ^2(\theta )-$$ $$4 \cos (2 \alpha )+4 \cos (2 \beta )-12$$ $$c=-16 \cos (\alpha ) \cos (2 \beta ) \cot (\theta ) \csc (\theta )+4 \sin ^2(\alpha ) \cos (2 \beta )+$$ $$\csc ^2(\theta ) (\cos (2 \alpha ) (\cos (2 \theta )+3)+9)+3 \cot ^2(\theta )-3$$

Applied to the case use by @Jean Marie $\alpha=\theta=\frac\pi 4 $ and $\beta=\tan ^{-1}(2))$, this leads to $$132 t^4-152 t^2+36=0$$ giving the four roots $$t=\pm \frac{1}{\sqrt{3}} \qquad \text{and} \qquad t=\pm \frac{3}{\sqrt{11}} $$ that is to say $$x=\pm \frac \pi 3\approx \pm1.04720 \qquad \text{and} \qquad x=\pm2 \tan ^{-1}\left(\frac{3}{\sqrt{11}}\right)\approx 1.47063$$ and then the corresponding $y$'s using $(3)$ or $(4)$.

Answer 2

$$\cos \alpha = \cos \theta \cos x + \sin \theta \sin x \cos y$$ $$\tan \beta = \frac{\cos x +\cos \theta \cos \alpha}{\sin \theta \sin x \sin y} $$ you set $\cos x=t;\;\sin x=u;\;\cos y=v;\;\sin y=w$ and get $$\cos \alpha = \cos \theta\, t + \sin \theta\, u v$$ $$\tan \beta = \frac{t +\cos \theta \cos \alpha}{\sin \theta\, u w} $$

add the conditions $$t^2+u^2=1;\;v^2+w^2=1$$

If you send the constant values I can try to solve it for you.

It is too complicated to solve it in general.

Answer 3

Not an answer, just illustrating what I said in my comment:

Answer 4

It's doubtful there are explicit formulas as you write them. Nevertheless, using tangent half angle formulas:

$$\cos x=\dfrac{1-t^2}{1+t^2}, \ \ \sin x = \dfrac{2t}{1+t^2}, \ \ \tan x=\dfrac{2t}{1-t^2}$$

$$\cos y=\dfrac{1-u^2}{1+u^2}, \ \ \sin y = \dfrac{2u}{1+u^2}, \ \ \tan y=\dfrac{2u}{1-u^2}$$

with $t=\tan \tfrac{x}{2} \ \text{and} \ \ u=\tan \tfrac{y}{2}$, you have a method of attack providing solutions in (most) trigonometric equations.

Indeed, with these formulas, the given system:

$$\begin{cases}\cos \alpha & =& \cos \theta \cos x + \sin \theta \sin x \cos y\\ \tan \beta & = & \dfrac{\cos x +\cos \theta \cos \alpha}{\sin \theta \sin x \sin y}\end{cases}$$

becomes, with evident notations:

$$\begin{cases}C_1 & = & C_3 \dfrac{1-t^2}{1+t^2}+S_3\dfrac{2t}{1+t^2} \dfrac{1-u^2}{1+u^2}\\ T_2 & = & \dfrac{\dfrac{1-t^2}{1+t^2} +C_3 C_1}{S_3 \dfrac{2t}{1+t^2} \dfrac{2u}{1+u^2}}\end{cases}$$

In this way, your issue is brought back to the intersection between 2 curves with (in general) 4th degree polynomial equations:

$$\begin{cases}S_3 2t(1-u^2)& = &(1+u^2)(C_1(1+t^2)-C_3(1-t^2)) \ \ \ & (a)\\4T_2S_3tu & = & (1+u^2)((1-t^2)+C_1C_3 (1+t^2)) \ \ \ & (b)\end{cases}\tag{1}$$

which can have at most $4 \times 4 = 16 $ real solutions $(t_k,u_k)$.

Here is a case with 4 solutions (corresponding to case $\alpha=\theta=\pi/4$ and $\beta=$arctan $2$): the curve in blue (resp. in red) has equation (a) (resp. (b)) in (1); ($t$ on abscissas and $u$ on the ordinates) giving, for the indicated root, $t=0.5782=\tan \dfrac{x}{2} \implies x=1.0485 $ radians and the same thing for $u=0.5164 =\tan \dfrac{y}{2} \implies y=0.9534 $ radians.

---

Edit: An afterthought after having seen the elegant solution by Claude Leibovici.

I didn't notice at first that equations (1) (a) and (b) are separable under the form:

$$\begin{cases}\dfrac{1-u^2}{1+u^2}& = &\dfrac{C_1(1+t^2)-C_3(1-t^2)}{S_3 2t} \ \ \ & (a)\\ \dfrac{2u}{1+u^2} & = & \dfrac{(1-t^2)+C_1C_3 (1+t^2)} {2T_2S_3t}\ \ \ & (b)\end{cases}\tag{2}$$

where we recognize in the two LHS $\cos y$ and $\sin y$ resp. Expressing that $\cos^2 y+\sin^2 y=1$ gives the same kind of fourth degree equation for $t$ as Claude has found...