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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 $\theta_2 + \theta_3$?

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By, for example, $theta2$, do you mean $\theta_2$ or $2\theta$? – Shaun Feb 9 '14 at 18:29
You might find this helpful :) – Shaun Feb 9 '14 at 18:31
no, it's theta2 because it's second theta of the manipulator, so it's only a name for it. Thanks for editing :) and for link. – user126448 Feb 9 '14 at 18:31
up vote 2 down vote accepted

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.

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I forgot to write in my post that those are degrees values, not radians. – user126448 Feb 9 '14 at 18:37
and there should be $-35.5$ instead of $35.5$, sorry. – user126448 Feb 9 '14 at 18:44
You still can't have the cosine of something being $45$, whether degrees or radians – Ross Millikan Feb 9 '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 9 '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 1:34

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