0
$\begingroup$

I have:

$$y(k)=a\ \sin(k+b)$$

I want this form "transform" into matrix(expose a and b), so it will looks like:

$y(k)= A \ [a \ \ b]^{T}$

where matrix $A$ does not contain $a$ and $b$.

where is known relation: $\sin(k+b)\sin(k)\cos(b)+\cos(k)\sin(b)$

  • 0
    And now, what we do now? We downvote and close this as too localized? not a real question? off topic?2011-05-07
  • 1
    @Fabian: I have tried to clean up, now Joker can clarify.2011-05-07
  • 1
    Going from function to matrix is a bit of a jump, no? What are you *really* trying to do?2011-05-07
  • 0
    @Moron: I'm curious...2011-05-07
  • 2
    @Fabian: I am not claiming this is a real question :-). Only that let us give Joker a chance before we close it soon.2011-05-07
  • 0
    why close? what I do wrong?2011-05-07
  • 0
    @Joker: what is expose? (I'm no native speaker)2011-05-07
  • 1
    You can try answering my question, first of all... and this other one: why do you need to do this?2011-05-07

1 Answers 1

1

What you are asking for, as I understand it, shall be something like trying to separate the unknown $k$ with the constant $a$ and $b$?

Just define vector function: $F(x)=[\sin(x) \cos(x)]^T$ ($X^T$ for the transpose of matrix $X$), and denote matrix $A = \begin{bmatrix}0&a\\a&0\end{bmatrix}$. Then we can express

$(k) = [F(b)^T]AF(k)$,

which shall be to your point, though not exactly in your form. Am I right?

Cheers!

Richard