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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
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1 Answer

up vote 2 down vote accepted

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.

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Wouldn't this give us the same equation we started off with? This includes theta and alpha so not sure what you mean by finding alpha by multiplying the first factor with z. –  JCs Sep 24 '11 at 21:13
    
@JCs: See my edit. –  Christian Blatter Sep 25 '11 at 7:31
    
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
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