Solve trigonometric equations with sin and cos

Question

I have two equations for two unknowns $u_k$ and $v_k$:

$\tan{u_1} \tan{u_k} + \cos{(v_1 - v_k)} = 0$

$\tan{u_2} \tan{u_k} + \cos{(v_2 - v_k)} = 0$

where $u_1$, $v_1$, $u_2$ and $v_2$ are parameters.

I managed to find a general solution for $v_k$ by expanding $\cos{(v_i - v_k)} = \cos{v_i} \cos{v_k} + \sin{v_i} \sin{v_k}$ and then dividing to obtain a general solution for $v_k$:

$\tan{v_k} = \frac{\tan{u_2} \cos{v_1} - \tan{u_1} \cos{v_2}}{\tan{u_1} \sin{v_2} - \tan{u_2} \sin{v_1}}$

However, I cannot find a general algebraic solution for $u_k$ not containing $v_k$. I have only obtained the following identities

$\tan{u_k} = \frac{\sin{(v_2 - v_1)}}{\tan{u_1} \sin{v_2} - \tan{u_2} \sin{v_1}} \cos{v_k}$

$\tan{u_k} = \frac{\tan{u_2} \cos{v_1} - \tan{u_1} \cos{v_2}}{\sin{(v_2 - v_1)}} \sin{v_k}$

but have been unable to get rid of the $v_k$ from either of them as I always start with both a sine and a cosine of it. Squaring the equations did not help, too.

Is there a way to get $u_k$ as a function of just $u_1$, $v_1$, $u_2$ and $v_2$?

Answer 1

Remember that if $\tan \theta=T$, then $\cos \theta =\sqrt{\frac{1}{1+T^2}}$ and $\sin \theta =\frac{T}{\sqrt{1+T^2}}.$ We are tacitly restricting $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$.

Now, inasmuch as you already have $\tan v_k$ in terms of the given parameters $u_1,u_2,v_1$, and $v_2$, then you can obtain both $\cos v_k$ and $\sin v_k$ in terms of the given parameters also. Let's call $C(u_1,u_2,v_1,v_2)=\cos v_k$ and $S(u_1,u_2,v_1,v_2)=\sin v_k$.

Substituting $C(u_1,u_2,v_1,v_2)$ and $S(u_1,u_2,v_1,v_2)$ into the expressions you have for $\tan u_k$ provide equations for $\tan u_k$ in terms of the parameters alone. Simply take the arctangent and you have $u_k$!

Answer 2

$$\tan{u_1} \tan{u_k} + \cos{(v_1 - v_k)} = 0\\ \tan{u_2} \tan{u_k} + \cos{(v_2 - v_k)} = 0$$ You reached: $$\tan{u_k} = \frac{\sin{(v_2 - v_1)}}{\tan{u_1} \sin{v_2} - \tan{u_2} \sin{v_1}} \cos{v_k},\tan{u_k} = \frac{\tan{u_2} \cos{v_1} - \tan{u_1} \cos{v_2}}{\sin{(v_2 - v_1)}} \sin{v_k}$$ Try calculating: $\sin^2v_k+\cos^2v_k=1$: $$\left(\frac{\tan u_k\sin(v_2-v_1)}{\tan{u_2} \cos{v_1} - \tan{u_1} \cos{v_2}}\right)^2+ \left(\frac{\tan u_k(\tan{u_1} \sin{v_2} - \tan{u_2} \sin{v_1})}{\sin{(v_2 - v_1)}}\right)^2=1$$ So: $$u_k=\arctan\left(\left(\frac{\sin(v_2-v_1)}{\tan{u_2} \cos{v_1} - \tan{u_1} \cos{v_2}}\right)^2+ \left(\frac{\tan{u_1} \sin{v_2} - \tan{u_2} \sin{v_1}}{\sin{(v_2 - v_1)}}\right)^2\right)^{-1/2}$$