Solve system of two trigonometric equations (related to flight mechanics)

Question

I have a constrained gimbal model which uses $\hat\phi$ and $\hat\theta$ to drive a camera. I derived following two equations connecting my gimbal to classical aircraft Euler angles (or Tait-Bryan angles whichever name you prefer) $\phi,\theta,\psi$

$\sin(\theta)=\sin(\hat\theta)*\cos(\hat\phi)$

$\tan(\psi)=\tan(\hat\theta)*\sin(\hat\phi)$

I want now to solve this system for $\hat\phi$ and $\hat\theta$ but my time at university is to long ago ;) Can you give me a hint or the solution? I already tried wolfram alpha but it only offers solutions which probably can be simplified a lot more.

Answer 1

Divide the first equation by $\sin(\hat\theta)$ and the second by $\tan(\hat\theta)$, square both of these & use pythagorus, we have \begin{eqnarray*} \left( \frac{\sin(\theta)}{\sin(\hat\theta)} \right)^2 = 1- \left( \frac{\tan(\psi)}{\tan(\hat\theta)} \right)^2. \end{eqnarray*} Now multiply by $\sin^2(\hat\theta)$ and use pythagorus \begin{eqnarray*} \sin^2(\theta) = 1-\cos^2(\hat\theta)- \cos^2(\hat\theta) \tan^2(\psi) \\ \end{eqnarray*} Rearrange a bit
\begin{eqnarray*} \cos^2(\hat\theta) = \frac{1-\sin^2(\theta)}{1+\tan^2(\psi)} \end{eqnarray*} square root to get \begin{eqnarray*} \cos(\hat\theta) = \pm \cos(\theta)\cos(\psi). \end{eqnarray*}

Answer 2

Hitn.

Calling

$$ u=\sin\hat\theta\\ v=\sin \hat\phi $$

we have

$$ \sin^2\theta = u^2(1-v^2)\\ \tan^2\psi = \frac{u^2}{1-u^2}v^2 $$

then

$$ 1-\frac{\sin^2\theta}{u^2}=\frac{1-u^2}{u^2}\tan^2\psi $$

so we find

$$ u^2 = \cos^2\psi(\sin^2\theta + \tan^2\psi) $$

etc.