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I have the following equations \begin{align*} R_1\cos(\omega T_1-\phi_{1})& =Q_1-R_2\cos(\omega T_2-\phi_{2})\\ R_1\sin(\omega T_1-\phi_{1})& =-Q_2-R_2\sin(\omega T_2-\phi_{2}). \end{align*}

From these equations how can I obtain the following solution $$\omega T_2=\pm \arccos\left(\frac{Q_1^2+Q_2^2-R_1^2+R_2^2}{2R_2\sqrt{Q_1^2+Q_1^2}}\right)-\arctan\left(\frac{Q_2}{Q_1}\right)+\phi_2+2k\pi .$$

Udichi
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    Welcome to Math.SE Udichi. Please write out your equations using Mathjax: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Rhys Hughes May 23 '18 at 12:39
  • What are the unknowns in your equation System? – Dr. Sonnhard Graubner May 23 '18 at 12:42
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    Define $\theta_i := \omega T_i-\phi_i$, so that the equations become $$\begin{align}R_1\cos \theta_1 = \phantom{-}Q_1-R_2 \cos\theta_2 \ R_1 ;\sin\theta_1 = -Q_1-R_2\sin\theta_2\end{align}$$ With less clutter, it's easier to see that you can eliminate $\theta_1$ by squaring the sides of each equation and adding. After some simplification, the resulting equation has the form $$a \sin\theta_2+b \cos\theta_2 + c = 0$$ which is subject to the approaches shown in this question. – Blue May 23 '18 at 13:17
  • What is the question ? –  May 23 '18 at 15:44
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    @Blue Thank you very much – Udichi May 24 '18 at 10:48

3 Answers3

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As mentioned in the comments by @Blue the problem can be reduced into $$ a \sin \theta + b \cos \theta +c = 0 $$

to be solved for $\theta$.

The way I have dealt with this equation in the past was to split the angle into two angles $\theta = \varphi + \psi$, and expand

$$\sin \theta = \sin(\varphi+\psi) = \cos\varphi \sin\psi + \sin\varphi \cos\psi$$ $$\cos \theta = \cos(\varphi+\psi) = \cos\varphi \cos \psi - \sin\varphi \sin\psi$$

Then my equation to be solved is

$$ \Bigl( \cos\varphi ( b \cos\psi + a \sin \psi) \Bigr) + \Bigl( \sin\varphi (a \cos\psi - b \sin\psi)+c \Bigr) =0 $$

and you can set each part equal to zero individually

$$\begin{aligned} \cos\varphi ( b \cos\psi + a \sin \psi) & = 0 \\ \sin\varphi (a \cos\psi - b \sin\psi)+c & = 0 \end{aligned} $$

The first one is solved by $\boxed{\psi = -\tan^{-1} \left( \frac{b}{a} \right) + k \pi }$ which transforms the second one into

$$ \sin\varphi \sqrt{a^2+b^2}+c = 0 $$

with solution $\boxed{ \varphi = \sin^{-1}\left( \frac{c}{\sqrt{a^2+b^2}} \right) +2 k \pi}$

So all together you have

$$ \boxed{ \color{blue}{ \theta = \sin^{-1}\left( \frac{c}{\sqrt{a^2+b^2}} \right)-\tan^{-1} \left( \frac{b}{a} \right) +2 k \pi} } $$

John Alexiou
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From $$a\sin\theta+b\cos\theta+c=0,$$

$$a^2\sin^2\theta=a^2(1-\cos^2\theta)=(b\cos\theta+c)^2$$

and

$$(a^2+b^2)\cos^2\theta+2bc\cos\theta+c^2-a^2=0.$$

You solve the quadratic equation for $\cos\theta$, and $\sin\theta$ follows.

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Another way to tackle the same problem of solving $a \sin \theta + b \cos \theta + c = 0$ is to use the tan-half-angle substitutions

$$\begin{aligned} t &= \tan \left( \frac{\theta}{2} \right)\, \rightarrow & \theta = 2 \arctan(t) \\ \sin(\theta) & = \frac{2 t}{1+t^2} \\ \cos(\theta) & = \frac{1-t^2}{1+t^2} \end{aligned}$$

The problem is transformed into a polynomial equation

$$\left. a \frac{2 t}{1+t^2} + b \frac{1-t^2}{1+t^2} + c = 0 \right\}\;\; t=\begin{cases} \frac{a+\sqrt{a^2+b^2-c^2}}{b-c} \\ \frac{a-\sqrt{a^2+b^2-c^2}}{b-c} \end{cases} $$

With the final and solution in terms of the angle

$$ \theta = \begin{cases} 2 \arctan\left( \frac{a+\sqrt{a^2+b^2-c^2}}{b-c} \right) \\ 2 \arctan\left( \frac{a-\sqrt{a^2+b^2-c^2}}{b-c} \right) \end{cases} $$

Example

$$ 5 \sin \theta + 2 \cos \theta - 3 = 0 $$

$$ t= \begin{cases} \frac{5}{2-(-3)} + \frac{ \sqrt{5^2+2^2-(-3)^2}}{2-(-3)} = 1.89443 \\ \frac{5}{2-(-3)} - \frac{ \sqrt{5^2+2^2-(-3)^2}}{2-(-3)} = 0.10557 \end{cases} $$

$$ \theta= \begin{cases} 2 \arctan(1.89443) = 2.170\,{\rm rad} \\ 2 \arctan(0.10557) = 0.2104\,{\rm rad} \end{cases} $$

Check

$$ \begin{cases} 5 \sin(2.170)+2 \cos(2.170) -3 \approx 0 \\ 5 \sin(0.2104) + 2\cos(0.2104) -3 \approx 0 \end{cases} $$

John Alexiou
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