# Solving trigonometric equation $\alpha = \arctan(o/q) - \arcsin(b/q)$

I have the following equation:

$$\alpha = \arctan\Big(\frac{a}{q}\Big) \ - \ \arcsin\Big(\frac{b}{q}\Big)$$

The values $$\alpha$$, $$a$$ and $$b$$ are known, the only missing value is $$q$$. So I need to solve the equation for $$q$$ but I have absolutely no idea how to do it or even where to start.

I also tried putting this equation into Wolfram Alpha, but even this didn't give a usable output.

I've tried something like this, but it doesn't seem that it would lead to the correct solution

$$\sin(\alpha) = \sin\bigg(\arctan\Big(\frac{a}{q}\Big)\bigg) - \frac{b}{q}$$

EDIT:

I've tried your hint @Kavi and I came up with the following:

\begin{align*} \tan(\alpha) &= \tan \Bigg(\arctan\Big(\frac{a}{q} - \arcsin\Big(\frac{b} {q}\Big)\Bigg) \\ &= \frac{\tan\Big(\arctan\big(\frac{a}{q}\big) - \tan\big(\arcsin\big(\frac{b}{q}\big)\big)\Big)}{1+\tan\Big(\arctan\big(\frac{a}{q}\big)\Big) \ \tan\Big(\arcsin\big(\frac{b}{q}\big)\Big)} \\ &= \frac{\frac{a}{q} - \frac{\frac{b}{q}}{\sqrt{1-\frac{b^2}{q^2}}}}{1+\frac{a}{q} \frac{\frac{b}{q}}{\sqrt{1-\frac{b^2}{q^2}}}} \end{align*}

Is this correct so far?

• $\sin\alpha = \frac{o \sqrt{1-\frac{b^2}{q^2}}-b}{q \sqrt{\frac{o^2}{q^2}+1}}$ – Cesareo Jul 12 '19 at 8:32
• Thanks for your comment. But how did you made this up and how could I proceed with it? – Codey Jul 12 '19 at 8:36
• $\sin(\arctan(t))=\frac{t}{\sqrt{t^2+1}}$. Now, squaring will lead to a montreous polynomial in $q$ – Claude Leibovici Jul 12 '19 at 8:42
• @ClaudeLeibovici the monstreous polynomial will still contain roots of $q$. No? – denklo Jul 12 '19 at 8:46
• @denklo. Yes, for sure ! One of the problems is that we shall need squaring which introduces extra roots. – Claude Leibovici Jul 12 '19 at 8:48

Hint: $$\tan (A-B)=\frac {\tan\, A-\tan \, B} {1+\tan \,A \tan \, B}$$. It is very easy to find $$q$$ once you apply this formula for $$\tan \, \alpha$$.
• Do you see any way to avoid a quartic polynomial in $q$ ? – Claude Leibovici Jul 12 '19 at 8:57