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How would you solve a system like this

$$ \left\{ \begin{aligned} 0&=E-\sin\left(\theta_1\right) + K \sin\left(\theta_2 - \theta_1\right)\\ 0&=E+\sin\left(\theta_2\right) - K \sin\left(\theta_2- \theta_1\right) \end{aligned} \right. $$

for $\theta_1$ and $\theta_2$?

I tried

Adding the two equations gives $$ \begin{aligned} 0=2E-\sin\left(\theta_1\right)+\sin\left(\theta_2\right)&\iff \sin\left(\theta_1\right) = 2E+\sin\left(\theta_2\right) \\ &\iff\sin\left(\theta_2\right)=-2E + \sin\left(\theta_1\right) \end{aligned} $$ Subtracting the first equation from the second gives $$ \begin{aligned} 0&=\sin\left(\theta_1\right) + \sin\left(\theta_2\right) -2K\sin\left(\theta_2- \theta_1\right)\\ &=\sin\left(\theta_1\right)+\sin\left(\theta_2\right)-2K\left(\sin\left(\theta_2\right)\cos\left(\theta_1\right)-\sin\left(\theta_1\right)\cos\left(\theta_2\right)\right)\\ &=\sin\left(\theta_1\right)\left(1+2K\cos\left(\theta_2\right)\right) + \sin\left(\theta_2\right)\left(1-2K\cos\left(\theta_1\right)\right)\\ \end{aligned} $$ From above, $$ \begin{aligned} 0&=\left(2E+\sin\left(\theta_2\right)\right)\left(1+2K\cos\left(\theta_2\right)\right) + \left(-2E + \sin\left(\theta_1\right)\right)\left(1-2K\cos\left(\theta_1\right)\right)\\ \iff 0&=2E+2EK\cos\left(\theta_2\right)+\sin\left(\theta_2\right)+K\sin\left(2\theta_2\right)\\ &-2E+2EK\cos\left(\theta_1\right)+\sin\left(\theta_1\right)-K\sin\left(2\theta_1\right) \end{aligned} $$ Moving all $\theta_2$ terms to the left hand side gives $$ \begin{aligned} -2EK\cos\left(\theta_2\right)-\sin\left(\theta_2\right)-K\sin\left(2\theta_2\right)&=2EK\cos\left(\theta_1\right)+\sin\left(\theta_1\right)-K\sin\left(2\theta_1\right) \\ \iff 2EK\cos\left(\pi-\theta_2\right)+\sin\left(-\theta_2\right)-K\sin\left(2\theta_2\right)&=2EK\cos\left(\theta_1\right)+\sin\left(\theta_1\right)-K\sin\left(2\theta_1\right)\\ \end{aligned} $$

But this doesn't seem to take me anywhere.

From problem formulation "I found that this problem is most easily done analytically and 'by hand'".

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  • $\begingroup$ Use $\cos x = \pm\sqrt{1 - \sin^2 x}$ to get rid of the cosines. $\endgroup$ – kennytm Apr 29 '15 at 8:19
  • $\begingroup$ How wold that help? If it helps, this will introduce certainly introduce false roots $\endgroup$ – user877329 Apr 29 '15 at 8:35
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May be, we could define $\theta_3=\theta_2-\theta_1$ and rewrite the equations as $$\left\{ \begin{aligned} 0&=E-\sin\left(\theta_1\right) + K \sin\left(\theta_3\right)\\ 0&=E+\sin\left(\theta_2\right) - K \sin\left(\theta_3\right) \end{aligned} \right.$$

So, we can solve for $\theta_1$ and $\theta_2$ as functions of $\theta_3,E,K$ $$\theta_1=\sin^{-1}\big(K\sin(\theta_3)+E\big)$$ $$\theta_2=\sin^{-1}\big(K\sin(\theta_3)-E\big)$$ which then gives $$\theta_3=\sin^{-1}\Big(K\sin(\theta_3)-E\Big)-\sin^{-1}\Big(K\sin(\theta_3)+E\Big)$$ $$\sin(\theta_3)=\sin\Big(\sin^{-1}\big(K\sin(\theta_3)-E\big)-\sin^{-1}\big(K\sin(\theta_3)+E\big)\Big)$$Now, setting $x=\sin(\theta_3)$ and expending, we can end with the equation $$x=(Kx-E) \sqrt{1-(E+K x)^2}-(K x+E)\sqrt{1-(E-K x)^2} $$ which is not the most pleasant but which could be solved using some numerical methods (the problem is so reduced to a single algebraic equation for a single unknwon).

Edit

In fact, this equation has potentially four roots which are given by $$x_{1,2}=\pm \frac{\sqrt{\frac{ \left(8 E^2+4\right) K^2-\sqrt{-\left(4 K^2+1\right)^2 \left(16 E^2 K^2-1\right)}+1}{K^4}}}{2 \sqrt{2}}$$ $$x_{3,4}=\pm \frac{\sqrt{\frac{\left(8 E^2+4\right) K^2+\sqrt{-\left(4 K^2-1\right)^2 \left(16 E^2 K^2-1\right)}-1}{K^4}}}{2 \sqrt{2}}$$ but some of them have been introduced by the successive squaring processes.

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