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Given the following two equations, where $c_1, c_2$ are independent constants

  1. $10(\cos\theta_1) + 10(\cos\theta_2) + 6(\cos\theta_3) = c_1$
  2. $10(\sin\theta_1) + 10(\sin\theta_2) + 6(\sin\theta_3) + 8= c_2$

Is it even possible to somehow simplify this system of equations to solve for $\theta_1, \theta_2, \theta_3$ ? I have tried using sum to product but it seems like a dead end.

Can anyone confirm that the only way to arrive at solutions for this is to use a optimization program?

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    $\begingroup$ don't you need at least $n$ equations to solve an $n$ variable system? $\endgroup$ Oct 5, 2015 at 2:16
  • $\begingroup$ Yes that is something I remember from classes in college linear algebra but is it applicable to systems of trig equations? $\endgroup$
    – jwj11iv
    Oct 5, 2015 at 2:23
  • $\begingroup$ Mathematica quits on me. I'm assuming this holds for trig equations too. In fact, I've seen systems with the correct number of trig equations that cannot be solved analytically... trig functions just work like that $\endgroup$ Oct 5, 2015 at 2:36
  • $\begingroup$ Where did this Q arise??? $\endgroup$ Oct 5, 2015 at 3:19
  • $\begingroup$ Rather late to join in to this conversation, but it doesn't matter what sort of equations you have. In general, it you have $n$ unknowns, you need $n$ equations to have a finite set of solutions. In this case, you have 3 unknowns but only two equations. The best you can do is solve for two of the variables in terms of the third, but that third variable will have a range of values each of which leads to a solution of this pair of equations. $\endgroup$ Nov 1, 2015 at 2:42

1 Answer 1

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Partial answer :Let $(\theta_1 +\theta_2)=m$ and $(\theta_1-\theta_2)/2=d.$ We have $\theta_1=m+d$ and $\theta_2=m-d$ from which $\cos \theta_1 +\cos \theta_2= 2(\cos m )(\cos d)$ and $\sin \theta_1+\sin \theta_2=2 (\sin m)(\cos d)$.So we have $$20(\cos m)\cos d=U$$ $$\text{and }20(\sin m)\cos d=V$$ where $$U=c_1-6\cos \theta_3$$ $$\text{and }V=c_2-8-6\sin \theta_3.$$ Now $U=V=0$ is possible for some $\theta_3$ iff $36=c_1^2+(8-c_2)^2$.In this case we must have $\cos d=0$ because $\cos m, \sin m$ cannot both be $0$.This means that $\theta_1$ and $\theta_2$ can be any values differing by an odd multiple of $\pi$, while $\cos \theta_3=c_1/6$ and $\sin \theta_3=(8-c_2)/6$....In general if $U$ and $V$ are any values for which $$(U-c_1)^2+(V-c_2+8)^2=36$$ then $\theta_3$ can exist. Now for $U,V$ not both $0$ we have $$20\ge|20 \cos d|=\sqrt{U^2+V^2}$$ which appears to place another constraint on $c_1,c_2$, And we have $$ \tan m=V/U \text{ or }\cot m=U/V.$$ Given allowable values for $c_1,c_2$ we can find values for $d,m$ from the above two equations, and obtain $\theta_1=m+d$ and $\theta_2=m-d$.

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