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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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    as$y$mmetric SVD?2011-09-25

1 Answers 1

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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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    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.2011-09-25