# Solving an equation with trig functions and two different angles

I am trying to solve this equation derived from matrix multiplication (where $a,b,c,d$ are constants):

$$-a \cos(\theta) \sin(\alpha)-b\sin(\theta) \sin(\alpha)+c\cos(\theta)\cos(\alpha)+d\sin(\theta)\cos(\alpha) = 0$$

The answer should be $\theta = (x_1+x_2)/2$ , $\alpha = (x_1-x_2)/2$ where $x_1 = \arctan((c-b)/(a+d))$ and $x_2 = \arctan((c+b)/(a-d))$

Here is what I have so far:

\begin{align} a(-\sin(\theta+\alpha)+\sin(\theta-\alpha))&+b(\cos(\theta-\alpha)+\cos(\theta+\alpha))+c(-\cos(\theta-\alpha) \\&+ \cos(\theta+\alpha))+d(\sin(\theta+\alpha)+\sin(\theta-\alpha))=0 \end{align}

I am stuck at this point though

Thanks

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These equations are hard to read. Please format them by enclosing them in dollar signs, using \sin, \cos, \arctan, \theta and \alpha and e.g. x_1 for $x_1$. – joriki Sep 24 '11 at 17:40
Can only read it with difficulty. Looks like a single equation, though. Very unlikely that it is enough to determine $\theta$ and $\alpha$. – André Nicolas Sep 24 '11 at 17:44
"derived from matrix multiplication" - what exactly were you trying to do that led you to this? A better solution might be found if you post the actual problem you're trying to solve. – J. M. Sep 24 '11 at 17:46
I am trying to solve an asymmetric SVD.. – JCs Sep 24 '11 at 17:50
asymmetric SVD? – user13838 Sep 25 '11 at 10:41

The given equation is equivalent to the matrix equation $$\left[\matrix{-\sin\alpha &\cos\alpha \cr}\right]\ \left[\matrix{a & b \cr c & d\cr}\right]\ \left[\matrix{\cos\theta \cr \sin\theta\cr}\right]\ =\ 0\ .$$ For any $\theta$ the product of the second and third factor gives a certain vector $v=\left[\matrix{v_1\cr v_2\cr}\right]$, and then it's easy to find an $\alpha$ such that the scalar product of $\ \left[\matrix{-\sin\alpha &\cos\alpha \cr}\right]$ with $v$ is $0$.
This shows that your equation does not determine $\alpha$ and $\theta$ (as already remarked in a comment by André Nicolas), but defines a certain dependency between these two variables: Given $\theta$, the value of $\alpha$ is determined up to multiples of $\pi$, and conversely.
You have to require that the second factor, the matrix is full rank otherwise you have all possible $\alpha$s as solutions. Since the product of second and the third would give zero already. – user13838 Sep 25 '11 at 10:39