Equation system

Given the following system of trigonometric equations

$ d_x=\sin{\alpha}\cdot\sin{\beta}\\ d_y=-\sin{\alpha}\cdot\cos{\beta}\\ d_z=\cos{\alpha}$

The unknowns are $\alpha, \beta$.

Solution set

It is easy to solve $\alpha$ with the last equation:

$\alpha \in \{\arccos{d_z} + 2\pi k\enspace|\enspace k \in \mathbb Z\} \cup \{2\pi - \arccos{d_z} + 2\pi k\enspace|\enspace k \in \mathbb Z\}$

This is a huge set of solutions, and I hope that the number of solutions can be reduced by regarding the first two equations and also the following knowledge.

Additional knowledge

$(d_x,d_y,d_z)$ forms a normal vector $d$, so the following is true:

  • $|d|=1$
  • $d_x,d_y,d_z\in[-1,1]$

My attempt

My attempt to reduce the solution set for $\alpha$ is the following:


Because $\cos{\beta} \in [-1,1]$, it must hold

$-1 \leq -\frac{d_y}{\sin{\alpha}} \leq 1$

Is this a correct assumption? If yes, I can rearrange it like

$ -1 \leq -\frac{d_y}{\sin{\alpha}} \leq 1 \quad \Leftrightarrow \quad -1 \geq -\frac{\sin{\alpha}}{d_y} \geq 1 $

$d_y$ can be negative and positive, so there two cases to consider:

$ d_y>0: \quad -d_y \geq -\sin{\alpha} \geq d_y \\ d_y<0: \quad -d_y \leq -\sin{\alpha} \leq d_y \\ $

But both cases are bad. $\geq$ is transitive, so $-d_y \geq d_y$ for positive $d_y$ doesn't makes sense.

Have you any other idea to reduce to solution set?

  • 1
    $\begingroup$ $tan(\beta)=-\frac{d_x}{d_y}$ if that helps. $\endgroup$
    – Paul
    Feb 25, 2016 at 9:56

2 Answers 2


Since $\alpha$ and $\beta$ only occur in trigonometric functions, you can always add $2\pi$ to either of them in any solution and obtain another solution. So there's nothing to be done about the "huge set of solutions", and no need to do anything about it. All you need to do is figure out the right quadrants for the angles.

Also, in this case, even apart from multiples of $2\pi$, both of your solutions for $\alpha$ lead to a solution for $\beta$. So there's nothing to reduce; you specified the correct set of solutions for $\alpha$, and each of these solutions corresponds to precisely one value of $\beta$ (given that $|d|=1$).


$$\alpha=\arctan\left(\frac{\sqrt{d_x^2+d_y^2}}{d_z}\right),\\ \beta=-\arctan\left(\frac{d_x}{d_y}\right),$$

Use the sign information to get the $\arctan$ in the proper quadrant (like the computer function $\text{atan2}$).

Then there is an extra solution

$$\alpha'=-\alpha,\\ \beta'=\beta+\pi.$$

  • $\begingroup$ Cool. Regarding to your second equation, can I say $\beta \in \{atan2(d_x,d_y)+2\pi k\enspace | \enspace k \in \mathbb Z\}$? $\endgroup$
    – Vertex
    Feb 25, 2016 at 10:33
  • $\begingroup$ @Vertex: you dropped the minus, as well as the $'$ solution. $\endgroup$
    – user65203
    Feb 25, 2016 at 10:42
  • $\begingroup$ You're right, sorry. $tan$ is $\pi$-periodic, now it makes sense. So hope, that $\beta \in \{-atan(d_x,d_y)+\pi k \enspace | \enspace k \in \mathbb Z\}$ is the right solution for $\beta$ $\endgroup$
    – Vertex
    Feb 25, 2016 at 10:48
  • $\begingroup$ @Vertex: yep, and choosing $\beta$ determines the sign of $\alpha$. $\endgroup$
    – user65203
    Feb 25, 2016 at 10:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.