I need to solve these two equations for two unknown angles for a robot arm I am trying to program.
- $\sin\theta_{1} + \sin\theta_{2} = c_{1}$
- $\cos\theta_1 + \cos\theta_{2} = c_{2}$
I multiplied equation 1 with $\cos \theta_{1}$, and equation 2 with $\sin \theta _{1}$, and subtracted both equations and used a trig rule to arrive at this but I am stuck. How do I go further to solve one of the variables?
$\sin({\theta_{2}-\theta_{1}}) = c_{1}\cos{\theta_{1}} – c_{2}\sin{\theta_{1}} $
Alternatively for sum to product I get
- $2\sin({(\theta_{1}+\theta_{2})/2})\cos({(\theta_{1}-\theta_{2})/2}) = c_{1}$
- $2\cos({(\theta_{1}+\theta_{2})/2})\cos({(\theta_{1}-\theta_{2})/2}) = c_{2}$
Adding them up, simplifying, squaring gives me
$\cos^2 (\theta_{1}-\theta_{2})/2) = (c_{1}/2)^2 + (c_{2}/2)^2$
How can I go further this route to solve for the angles?
Best Answer
$\def\t#1{\theta_{#1}}$
\begin{align} \sin\t1+\sin\t2&=c_1 \tag{1}\label{1} \\ \cos\t1+\cos\t2&=c_2 \tag{2}\label{2} \\ \sin(\t2-\t1) &= c_1\cos\t1 - c_2\sin\t1 \tag{3}\label{3} \end{align}
The sum of squares of \eqref{2} and \eqref{1} gives
\begin{align} \cos^2\t1&+2\,\cos\t1\cos\t2+\cos^2\t2 + \sin^2\t1+2\,\sin\t1\sin\t2+\sin^2\t2 = c_1^2+c_2^2, \\ \cos\t1\cos\t2&+\sin\t1\sin\t2 = \cos(\t2-\t1) =\tfrac12(c_1^2+c_2^2)-1 . \end{align}
hence
\begin{align} \sin(\t2-\t1)&=\sqrt{1- \left( \tfrac12(c_1^2+c_2^2)-1 \right)^2} \end{align}
Next, $\eqref{1}\times \sin\t1+\eqref{2}\times \cos\t1$ results in \begin{align} 1+\cos(\t2-\t1) &= c_1\,\sin\t1+c_2\,\cos\t1 \tag{4}\label{4} \end{align}
Equations \eqref{4} and \eqref{3} present a $2\times2$ linear system
\begin{align} \begin{bmatrix} \phantom{-}c_1&c_2\\ -c_2&c_1 \end{bmatrix} \begin{bmatrix} \sin\t1\\ \cos\t1 \end{bmatrix} &= \begin{bmatrix} \tfrac12(c_1^2+c_2^2) \\ \sqrt{1- \left( \tfrac12(c_1^2+c_2^2)-1 \right)^2} \end{bmatrix} . \tag{5}\label{5} \end{align}
Unless $c_1,c_2$ are both zero (hence $\t2 = \t1-\pi$), solution to \eqref{5} provides us with $\sin\t1$ and $\cos\t1$ (check that $\sin^2\t1 + \cos^2\t1\equiv1$). Together with \eqref{1} and \eqref{2} we can get the $\sin$ and $\cos$ of the other angle. Also, hence the system is symmetric, the solution is either $\{\t1,\t2\}$ or $\{\t2,\t1\}$.