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I have a problem, I'm trying to solve inverse kinematics problem, but I have hit the wall. I have 2 equation for $$\cos(\theta_2) \cos(\theta_3) - \sin(\theta_2) \sin(\theta_3) = 45.7518$$ and for $$\sin(\theta_3) \cos(\theta_2) + \sin (\theta_2) \cos(\theta_3) = -32.7842$$

I know that $\theta_2 + \theta_3 = -35.5$, $\theta_2 = 30$ and $\theta_3=-65.5$ Unfortunately I have no idea how can I solve it to get those answers. It probably has something to do with trigonometry. I hope someone will help me. And I don't even know if I have given enough information to solve it.

All value here are in degrees.

I have another question - let's say for a moment that this $45.7518$ is the answer to first equation, now if I call a function $acos(45.7518)$ I will get the $\theta_2 + \theta_3$? And also $asin(-32.7842)$ to get value of $\theta_2 + \theta_3$?

user126448
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Do you recognize the angle-sum equations? Your first equation says $\cos(\theta_2+\theta_3)=45.7518$ This will never be true because the cosine is less than $1$ in absolute value. Similarly, the left side of the second is $\sin(\theta_2+\theta_3)$ and again the right side is too big. The three equations after "I know that" are inconsistent. Please review the problem and get it right.

Ross Millikan
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  • I forgot to write in my post that those are degrees values, not radians. – user126448 Feb 09 '14 at 18:37
  • and there should be $-35.5$ instead of $35.5$, sorry. – user126448 Feb 09 '14 at 18:44
  • You still can't have the cosine of something being $45$, whether degrees or radians – Ross Millikan Feb 09 '14 at 18:44
  • I do not want to argue because clearly you are somewhat of an expert, but I don't think it is wrong. I used acos function of this value and it works and it satisfies the condition of the function after I convert the value to radians ($0.7985$) – user126448 Feb 09 '14 at 18:56
  • It is the angle, like $\theta_2$, that is in degrees or radians. The value of $\cos x$ is always between $-1$ and $1$, regardless of the units of the angle. $\arccos (45.7518)$ should give an error unless you are working the complex numbers. – Ross Millikan Feb 10 '14 at 01:34
  • Well I am using java environment to calculate those things, so I used first function $Math.toRadians(45.7518)$ which gave me $0.7985$ and then $Math.acos(0.7985)$, so it was between -1 and 1;p – user126448 Feb 10 '14 at 02:04
  • The point is that $45.7518$ is not in degrees. The only way I can imagine you got here is using a small angle approximation, where you used $\sin \theta \approx \theta$. Here if you measure $\theta$ in degrees the relation is $\sin \theta \approx \pi \theta/180$ and you are making a compensating mistake here. – Ross Millikan Feb 10 '14 at 02:15
  • Okay then, it's an approximation, but the point is if it is possible to get the approximated values of $\theta_2$ and $\theta_3$, using those equations, or any other way. Nevermind the results. – user126448 Feb 10 '14 at 08:06
  • As they appear in both equations in the combination $\theta_2+\theta_3$, no. You can determine the sum of the two, but not each individually. Having the two equations does resolve the ambiguity of the sign of $\theta_2+\theta_3$-the second tells you it is negative, so the sum is in the fourth quadrant. – Ross Millikan Feb 10 '14 at 15:10